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 long long int M = 1e3 + 5, LG = 26, SM = 800, inf = 1e9 + 1;
long long int mod = 95542721;
long long int dp[M][M];
long long int p9[M], p10[M];
long long int n, k;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k >> mod;
  p9[0] = p10[0] = 1;
  for (int i = 1; i < M; i++) {
    p9[i] = (p9[i - 1] * 10) % mod;
    p10[i] = (p10[i - 1] * 10) % k;
  }
  for (int i = 1; i < 10; i++) dp[0][i % k]++;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < 10; j++) {
      if (i == n - 1 && j == 0) continue;
      for (int t = 1; t < k; t++) {
        dp[i][(t + j * p10[i]) % k] += dp[i - 1][t];
        dp[i][(t + j * p10[i]) % k] %= mod;
      }
      if (j != 0) dp[i][(j * p10[i]) % k]++;
      dp[i][(j * p10[i]) % k] %= mod;
    }
  }
  long long int ans = 0;
  for (int i = 0; i < n - 1; i++) {
    ans += dp[i][0] * p9[n - i - 2] * 9;
    ans %= mod;
  }
  ans += dp[n - 1][0];
  ans %= mod;
  cout << ans << '\n';
}
