Codeforces Round #360 (Div. 2)

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Arya has n opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win, but implementing this plan requires presence of them all. That means if one day at least one of Arya's opponents is absent at the school, then Arya will beat all present opponents. Otherwise, if all opponents are present, then they will beat Arya.

For each opponent Arya knows his schedule — whether or not he is going to present on each particular day. Tell him the maximum number of consecutive days that he will beat all present opponents.

Note, that if some day there are no opponents present, Arya still considers he beats all the present opponents.

Input

The first line of the input contains two integers n and d (1 ≤ n, d ≤ 100) — the number of opponents and the number of days, respectively.

The i-th of the following d lines contains a string of length n consisting of characters '0' and '1'. The j-th character of this string is '0' if the j-th opponent is going to be absent on the i-th day.

Output

Print the only integer — the maximum number of consecutive days that Arya will beat all present opponents.

Examples

input

2 21000

output

input

4 10100

output

input

4 511011111011010111111

output

Note

In the first and the second samples, Arya will beat all present opponents each of the d days.

In the third sample, Arya will beat his opponents on days 1, 3 and 4 and his opponents will beat him on days 2 and 5. Thus, the maximum number of consecutive winning days is 2, which happens on days 3 and 4.

每次输入字符串，如果字符串中有0，就代表Arya赢了，计算Arya最大连续赢几局，手速依然太慢啊，加强锻炼要，代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int main(){    char str[105];    int n,d;    while(~scanf("%d%d",&n,&d))    {        int sum=0,flag=0;        int maxn=0;        for(int i=0;i<d;i++)        {            scanf("%s",str);            for(int i=0;i<strlen(str);i++)            {                if(str[i]=='0')                {                    flag=1;                    break;                }            }            if(flag)                sum++;            else                sum=0;            maxn=max(sum,maxn);            flag=0;        }        printf("%d\n",maxn);    }    return 0;}

Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.

Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.

Now Pari asks you to write a program that gets a huge integer n from the input and tells what is the n-th even-length positive palindrome number?

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10^100 000).

Output

Print the n-th even-length palindrome number.

Examples

input

output

input

output

Note

The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.

直接输出字符串和其反串：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;char str[100005],s[100005];int main(){    while(~scanf("%s",str))    {        int j=0;        for(int i=strlen(str)-1;i>=0;i--)            s[j++]=str[i];        s[j]='\0';        printf("%s%s\n",str,s);    }    return 0;}

Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.

Suppose the graph G is given. Subset A of its vertices is called a vertex cover of this graph, if for each edge uv there is at least one endpoint of it in this set, i.e. $\text{[math]}$ or $\text{[math]}$ (or both).

Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.

They have agreed to give you their graph and you need to find two disjoint subsets of its vertices A and B, such that both A and B are vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — the number of vertices and the number of edges in the prize graph, respectively.

Each of the next m lines contains a pair of integers u_i and v_i (1 ≤ u_i, v_i ≤ n), denoting an undirected edge between u_i and v_i. It's guaranteed the graph won't contain any self-loops or multiple edges.

Output

If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).

If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer k denoting the number of vertices in that vertex cover, and the second line contains kintegers — the indices of vertices. Note that because of m ≥ 1, vertex cover cannot be empty.

Examples

input

4 21 22 3

output

12 21 3

input

3 31 22 31 3

output

-1

Note

In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to Pari and keep vertex number 4 for yourself (or give it someone, if you wish).

In the second sample, there is no way to satisfy both Pari and Arya.

DFS二分图染色，不会啊，附大牛博客
二分图染色

Today Pari and Arya are playing a game called Remainders.

Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value $\text{[math]}$ . There are n ancient numbers c₁, c₂, ..., c_n and Pari has to tell Arya $\text{[math]}$ if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value $\text{[math]}$ for any positive integer x?

Note, that $\text{[math]}$ means the remainder of x after dividing it by y.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.

The second line contains n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 1 000 000).

Output

Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x, or "No" (without quotes) otherwise.

Examples

input

4 52 3 5 12

output

Yes

input

2 72 3

output

No

Note

In the first sample, Arya can understand $\text{[math]}$ because 5 is one of the ancient numbers.

In the second sample, Arya can't be sure what $\text{[math]}$ is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.

中国剩余定理，见中国剩余定理

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;long long gcd(long long a,long long b){    if(b==0) return a;    return gcd(b,a%b);}int main(){    long long n,k;    while(~scanf("%lld%lld",&n, &k))    {        long long x,sum =  1;        for (int i=0;i<n;i++)        {            scanf("%lld",&x);            sum = x / gcd(x, sum)*sum % k;//求最小公倍数        }        if(sum==0)            cout<<"Yes"<<endl;        else            cout<<"No"<<endl;    }    return 0;}

看见了一些大牛的代码，看见有

std::__gcd(c, k);

貌似是#include<algorithm>头文件里面的库函数，这里记一下，以后可以直接用了。

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