#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int n, m, mod;
const int N = 1e6 + 10;
int ne[N];    // kmp的ne数组，针对模式串s
char s[N];    // 模式串s
int f[N][22]; // dp数组

//原始版本，就是设计密码那道题的代码，可以过4个点，剩余6个点TLE
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;
    }

    //普通dp
    f[0][0] = 1;
    for (int k = 0; k < n; k++)
        for (int i = 0; i < m; i++)
            for (char c = '0'; c <= '9'; c++) {
                int j = i;
                while (j && s[j + 1] != c) j = ne[j];
                if (s[j + 1] == c) j++;
                //在kmp过程中进行判断,不能命中m个长度，需要避让
                if (j < m) f[k + 1][j] = (f[k + 1][j] + f[k][i]) % mod;
            }

    int res = 0;
    for (int i = 0; i < m; i++) res = (res + f[n][i]) % mod;

    printf("%d\n", res);
    return 0;
}