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: 
    * <image>; 
    * 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.
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long infLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1000 + 5;
const int maxk = 100 + 5;
int n, k, m;
int f[maxn][maxk][2];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  f[0][0][0] = 1;
  for (int i = 1, p = 1; i <= n; ++i, p = p * 10 % k) {
    for (int j = 0; j < (k); ++j)
      for (int d = (i == n); d <= (9); ++d) {
        int t = (j + (long long)d * p) % k;
        if (!t && d) {
          f[i][t][1] = (f[i][t][1] + f[i - 1][j][0]) % m;
        } else {
          f[i][t][0] = (f[i][t][0] + f[i - 1][j][0]) % m;
        }
        f[i][t][1] = (f[i][t][1] + f[i - 1][j][1]) % m;
      }
  }
  int res = 0;
  for (int j = 0; j < (k); ++j) res = (res + f[n][j][1]) % m;
  printf("%d\n", res);
  return 0;
}
