897B. Chtholly's request

— 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

Copy

2 100

Output

33

Input

Copy

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：求出前k个zcy数（长度为偶数的回文字符串）和对p求模

思路：构造zcy数

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N =1e5+10;
ll per[N<<1];
int k,p;
ll get(int x){
	ll ans=(ll)x;
	while(x>0){
		ans=ans*10+x%10;
		x/=10;
	}
	return ans;
}
void init(){
	per[0]=0;
	for(int i=1;i<=N;i++)
	   per[i]=(get(i)+per[i-1])%p;
}
int main(){
    cin>>k>>p;
    init();
	cout<<per[k]<<endl;
	return 0;
}