Codeforces Round #449 (Div. 2) B. 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 is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.

Given integers k and p, 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, 

.

比赛的时候少看到一个条件。。以为12321这样的数也是zcy所以没做出来。。。。第二天早上才看到。。。。

第几个zcy数 就是 由第几个数进行回文复制的。。 比如第12个zsy数就是 1221， 第23个就是2332.。。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <vector>
#include <array>
#include <algorithm>
using namespace std;
using ll = long long;
ll k, p;
ll get(ll i) {
    ll res = i;
    while (i) {
        res = res * 10 + i % 10;
        i /= 10;
    }
    return res;
}

int main() {
    //freopen("in.txt", "r", stdin);
    cin >> k >> p;
    ll sum = 0;
    while (k) {
        sum += get(k);
        //cout << get(k) << endl;
        sum %= p;
        k--;
    }
    cout << sum << endl;
}