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 mem[1001][100][2];
long long rm[10][1001];
long long n, k, m;
long long dp(long long cur, long long r, long long t) {
  if (cur >= n) return t;
  long long &ret = mem[cur][r][t];
  if (ret != -1) return ret;
  ret = 0;
  long long s = (cur + 1 == n);
  for (long long i = s; i <= 9; i++) {
    if ((rm[i][cur] + r) % k == 0 && i)
      ret += dp(cur + 1, (rm[i][cur] + r) % k, 1);
    else
      ret += dp(cur + 1, (rm[i][cur] + r) % k, t);
    ret %= m;
  }
  return ret;
}
void calc() {
  for (long long i = 0; i <= 9; i++) {
    rm[i][0] = i % k;
    for (long long j = 1; j <= n - 1; j++) rm[i][j] = (rm[i][j - 1] * 10) % k;
  }
  return;
}
int main() {
  cin >> n >> k >> m;
  calc();
  memset(mem, -1, sizeof(mem));
  cout << dp(0, 0, 0) << endl;
  return 0;
}
