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 dp[1005][101][2];
long long pw[1005];
int n, m, k;
void Set() {
  pw[0] = 1;
  for (int i = 1; i < 1001; i++) {
    pw[i] = pw[i - 1] * 10;
    pw[i] %= k;
  }
}
long long cal(int cp, int cm, int cf) {
  if (dp[cp][cm][cf] != -1) return dp[cp][cm][cf];
  if (cp == 0 && cm == 0) return 1;
  if (cp == 0) return 0;
  dp[cp][cm][cf] = 0;
  int st = 0;
  if (cp == 1) st = 1;
  for (int i = st; i < 10; i++) {
    dp[cp][cm][cf] +=
        cal(cp - 1, cm == 0 && cf ? 0 : ((cm + i * pw[n - cp]) % k),
            i > 0 ? 1 : cf);
    dp[cp][cm][cf] %= m;
  }
  return dp[cp][cm][cf];
}
int main() {
  cin >> n >> k >> m;
  Set();
  memset((dp), -1, sizeof((dp)));
  cout << cal(n, 0, 0) << endl;
  return 0;
}
