codeforce 897B

B. Chtholly's request

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

— Thanks a lot for today.

— I experienced so many great things.

— You gave me memories like dreams... But I have to leave now...

— One last request, can you...

— Help me solve a Codeforces problem?

— ......

— What?

Chtholly has been thinking about a problem for days:

If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number ispalindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and12321 is not.

Given integers k andp, calculate the sum of the k smallest zcy numbers and output this sum modulo p.

Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help!

Input

The first line contains two integers k and p (1 ≤ k ≤ 10⁵, 1 ≤ p ≤ 10⁹).

Output

Output single integer — answer to the problem.

Examples

Input

2 100

Output

33

Input

5 30

Output

15

Note

In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.

In the second example, .

题意：如果一个数的位数是偶数并且正反读一样的话，那么就称这个数为zcy的数，给你k和p问你从最小的zcy数开始，连续k个数的和对p取余的结果；

思路：直接模拟就好了，首先要找到对应关系，1~11,2~22，。。。。。10~1001，所以第几小的数就是第几的回文串，所以相加取余即可；

下面附上我的代码：

/*1~11  2~22  3~33  10~1001  15~1551*/#include<bits/stdc++.h>using namespace std;typedef long long LL;LL a[20];LL k,p;LL decimal(LL m)//求位数{    int res=0;    while(m)    {        m/=10;        res++;    }    return res;}LL solve(LL n){    int p;    int k=decimal(n);    p=2*k-1;    LL ans=0;    while(n)    {        a[p-k]=a[k]=n%10;//直接赋成回文串        n/=10;        k++;    }   //for(int i=0;i<=p;i++)     //   printf("%d",a[i]);    //puts("");    for(int i=0;i<=p;i++)        ans=ans*10+a[i];    //printf("%d ll%d\n",p,ans);    return ans;}int main(){    while(cin>>k>>p)    {        LL s=0;        for(int i=1;i<=k;i++)                s=(s+solve(i))%p;//求和取余即可        printf("%lld\n",s);    }    return 0;}