Chtholly's request （思维）

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 and p, calculate the sum of the k smallest zcy numbers and output this sum modulop.

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 andp (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 is22.

In the second example, .

题意：前K个长度为偶数的回文数相加%p；

思路：长度为偶数，那么每个数对称一下都是符合要求的回文数

AC代码：

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<queue>using namespace std;int k,p;long long zcy[100005];void init(){    int cnt=0;    for(int i=1;i<=100000;i++)    {        long long tmp=i;        int p=i;        while(p){            tmp=tmp*10+p%10;            p/=10;        }        zcy[++cnt]=tmp;    }}int main(){    init();    while(~scanf("%d%d",&k,&p))    {        long long sum=0;        for(int i=1;i<=k;i++){            sum+=zcy[i];            sum%=p;        }        printf("%lld\n",sum);    }    return 0;}