The Maths Lecture

Amr doesn’t like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn’t.

First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:

• Decimal representation of x (without leading zeroes) consists of exactly n digits;
• There exists some integer y > 0 such that:
  • ;
  • decimal representation of y is a suffix of decimal representation of x.

As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.

Can you help Amr escape this embarrassing situation?

Input

Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 10⁹).

Output

Print the required number modulo m.

Examples

input

1 2 1000

output

4

input

2 2 1000

output

45

input

5 3 1103

output

590

Note

A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.

Solution:

#include <cstring>
#include <vector>
#include  <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

const int N = 2010;

int f[N][N][2][2];

int main() {
int n, k, m;
cin >> n >> k >> m;
memset(f, 0, sizeof(f));
f[n - 1][0][0][0] = 1 % m;
int z = 1 % k;
int ans = 0;
for (int i = n - 1; i >= 0; i--) {
for (int r = 0; r < k; r++) {
      for (int w = 0; w < 2; w++) {
        for (int q = 0; q < 2; q++) {
          int ft = f[i][r][w][q];
          if (ft == 0) {
            continue;
          }
          for (int j = 0; j < 10; j++) {
            int nr = (r + z * j) % k;
            int nq = (q | (j != 0));
            int nw = (w | (nr == 0 && nq == 1));
            if (i == 0) {
              if (n == 1 || j != 0) {
                if (nw == 1) {
                  ans += ft;
                  if (ans >= m) ans -= m;
}
}
} else {
f[i - 1][nr][nw][nq] += ft;
if (f[i - 1][nr][nw][nq] >= m) f[i - 1][nr][nw][nq] -= m;
}
}
}
}
}
z = z * 10 % k;
}
printf("%d\n", ans % m);
return 0;
}