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;
long long M;
long long add(long long a, long long b) { return (a + b) % M; }
const long long N = 1001;
const long long K = 101;
long long dp[N][K][2][2];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k >> M;
  dp[0][0][0][0] = 1;
  long long cp = 1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < k; j++) {
      for (long long w = 0; w < 2; w++) {
        for (long long bz = 0; bz < 2; bz++) {
          for (long long dg = 0; dg < 10; dg++) {
            if (dg == 0 && i + 1 == n) continue;
            long long nw = j + cp * dg;
            nw %= k;
            long long cz = (bz | (dg > 0));
            long long cw = (cz && nw == 0);
            dp[i + 1][nw][w | (cw)][cz] =
                add(dp[i + 1][nw][w | (cw)][cz], dp[i][j][w][bz]);
          }
        }
      }
    }
    cp *= 10;
    cp %= k;
  }
  long long ans = 0;
  for (long long j = 0; j < k; j++) {
    ans = add(ans, dp[n][j][1][1]);
  }
  cout << ans << endl;
}
