An abandoned sentiment from past

题目链接：http://codeforces.com/problemset/problem/814/A

A. An abandoned sentiment from past

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.

To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.

Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length kequals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.

If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.

Input

The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence aand b respectively.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.

The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.

Input guarantees that apart from 0, no integer occurs in a and b more than once in total.

Output

Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.

Examples

input

4 211 0 0 145 4

output

Yes

input

6 12 3 0 8 9 105

output

No

input

4 18 94 0 489

output

Yes

input

7 70 0 0 0 0 0 01 2 3 4 5 6 7

output

Yes

题目大意：Hitagi遇到麻烦，但是Kaiki有一个方法可以救Kaiki。解救方法为：输入两个整数序列a,b，系列a中‘0’的个数等于序列b的长度。用序列b中的数去代替序列a中的‘0’。但序列b中的数每个数只能使用一次。如果替换得到的系列a不是一个递增的序列，则Kaiki成功被解救，输出“Yes”，否则输出“No”。

解题思路：定义数组a、b分别存放序列a和序列b，将序列b进行全排列得许多序列。然后用每一种得到的序列去替换序列a中的‘0’，然后判断替换后的序列a是否为递增的序列。如果是输出“Yes”，否则输出“No”。这里的全排列需要用到next_permutataion函数。

代码：

#include<iostream>#include<algorithm>using namespace std;int main(){    int a[205],b[205],z[205],n,k,s,j;    while(cin>>n>>k)    {        for(int i=0;i<=n-1;i++)            cin>>a[i];        for(int i=0;i<=k-1;i++)            cin>>b[i];        sort(b,b+k);      //用next_permutation函数全排列数组时的前提是该数组是递增的顺序                          s=0;              //s只是一个中间变量，用来标记        j=0;        do        {            for(int i=0;i<=n-1;i++)    //该for循环的目的是：将a中不为0的数用b中的数来替换            {                if(a[i]!=0)                    z[i]=a[i];                else                {                    z[i]=b[j];                    j++;                }            }            for(int i=0;i<=n-2;i++)    //该循环的目的是：判断替换后的数组a书否为递增的数组            {                if(z[i]>=z[i+1])       //只需要判断数组a中后一个数是否比前一个数小，若果有后一个数比前一个数  小于或等于  的情况，则数组a一定不是递增的数组                {                    s=1;                    cout<<"Yes"<<endl;                    break;                }            }            if(s==1)                break;        }while(next_permutation(b,b+k));  //全排列函数        if(s==0)            cout<<"No"<<endl;    }    return 0;}