#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 25;

int n, m, mod;
char s[N];
int ne[N];
int a[N][N];

//矩阵乘法
void mul(int a[][N], int b[][N], int c[][N]) {
    int t[N][N] = {0};
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++)
            for (int k = 0; k < N; k++)
                t[i][j] = (t[i][j] + (LL)(a[i][k] * b[k][j]) % mod) % mod;
    }
    memcpy(c, t, sizeof t);
}

int main() {
    cin >> n >> m >> mod;
    cin >> s + 1;

    // kmp
    for (int i = 2, j = 0; i <= m; i++) {
        while (j && s[j + 1] != s[i]) j = ne[j];
        if (s[j + 1] == s[i]) j++;
        ne[i] = j;
    }

    // 初始化A[i][j]
    for (int i = 0; i < m; i++)
        for (int c = '0'; c <= '9'; c++) {
            int j = i;
            while (j && s[j + 1] != c) j = ne[j];
            if (s[j + 1] == c) j++;
            if (j < m) a[i][j]++;
        }

    // f[0][0]=1 base case
    int f[N][N] = {1};
    //矩阵快速幂
    for (int i = n; i; i >>= 1) {
        if (i & 1) mul(f, a, f); // f:基底 a:需要计算快速幂的常数矩阵 f:结果存储
        mul(a, a, a);            //倍增a
    }

    int res = 0;
    for (int i = 0; i < m; i++) res = (res + f[0][i]) % mod;
    printf("%d\n", res);
    return 0;
}