codeforces 897 B. Chtholly's request【构造回文数+递推】

B. Chtholly’s request

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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, .


题意： 规定zcy number的回文串的长度必须是偶数的，问你前k个zcy number的和模上p的值

分析： 可以直接构造zcy number回文串，递推求得1e5内的

参考代码

#include<bits/stdc++.h>

#define ll long long
using namespace std;
const int N = 1e5 + 10;

ll pre[N<<1];
ll n,p;
ll get(int x) {
ll res = (ll)x;
while(x > 0) {
res = res*10 + x%10;
x /= 10;
}
return res;
}
void init() {
for(int i = 1;i <= N;i++) {
pre[i] = (get(i) + pre[i-1])%p;
}
}

int main(){
ios_base::sync_with_stdio(0);
cin>>n>>p;
init();
cout<<pre
<<endl;
return 0;
}

如有错误或遗漏，请私聊下UP，thx