'''
获取KMP next数组, next[i]表示的是以i-1位置结尾的子串的最长公共前后缀长度
'''
from typing import List
def getKmpNext(data : List) -> List[int]:
    next = [0 for _ in range(len(data))]
    next[0] = -1
    j, k = 0, -1

    while j < len(data)-1:
        if k == -1 or data[j] == data[k]:
            j, k = j+1, k+1
            next[j] = k
        else:
            k = next[k]

    return next


def matrix_mul(A, B, mod):
    ans = [[0] * len(B[0]) for _ in range(len(A))]
    for i in range(len(A)):
        for j in range(len(B[0])):
            acc_val = 0
            for k in range(len(A[i])):
                acc_val += A[i][k] * B[k][j]

            ans[i][j] = acc_val % mod if mod is not None else acc_val

    return ans

def matrix_pow(A, k, mod):
    if k == 1:
        return A
    if k == 2:
        return matrix_mul(A, A, mod)

    t = matrix_pow(A, k >> 1, mod)
    ans = matrix_mul(A, matrix_mul(t, t, mod), mod) if k % 2 == 1 else matrix_mul(t, t, mod)
    #print(f'k = {k}, ans = {ans}')
    return ans



n, m, mod = map(int, input().split())
T = input()

A = [[0] * m for _ in range(m)]
D = getKmpNext(T)
for j in range(m):
    for ch in '0123456789':
        if ch == T[j]:
            if j+1 < m:
                A[j+1][j] += 1
        else:
            k = j
            while True:
                if k == -1 or T[k] == ch:
                    A[k+1][j] += 1
                    break
                else:
                    k = D[k]

if n == 1:
    if m > 1:
        print(10)
    else:
        print(9)
else:
    AA = matrix_pow(A, n-1, mod)
    if m == 1:
        ans = (AA[0][0] * 9) % mod
        print(ans)
    else:
        ans = 0
        for line in AA:
            ans += 9 * line[0] + line[1]
            ans %= mod

        print(ans)