Problem:
Score : 400 points

Problem Statement

Given is a string S. Each character in S is either a digit (0, ..., 9) or ?.

Among the integers obtained by replacing each occurrence of ? with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.

Since the answer can be enormous, print the count modulo 10^9+7.

Constraints

S is a string consisting of digits (0, ..., 9) and ?.

1 \leq |S| \leq 10^5

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of integers satisfying the condition, modulo 10^9+7.

Sample Input 1

??2??5

Sample Output 1

768

For example, 482305, 002865, and 972665 satisfy the condition.

Sample Input 2

?44

Sample Output 2

1

Only 044 satisfies the condition.

Sample Input 3

7?4

Sample Output 3

0

We may not be able to produce an integer satisfying the condition.

Sample Input 4

?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???

Sample Output 4

153716888

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

import sys
sys.setrecursionlimit(10**7)
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return sys.stdin.readline().strip()
INF = 10 ** 18
MOD = 10 ** 9 + 7

debug = True
debug = False
 
def dprint(*objects):
    if debug == True:
        print(*objects)

def solve():
    S = SI()
    N = len(S)
    S = reversed(S)
    dp = [[0 for j in range(13)] for i in range(N)]
 
    for i, es in enumerate(S):
        c = es
        if i == 0:
            if c == '?':
                for j in range(10):
                    dp[i][j] = 1
            else:
                dp[i][int(c)] = 1
        else:
            for k in range(13):
                if c == '?':
                    for j in range(10):
                        amari = (j * pow(10, i, 13)) % 13
                        dp[i][k] = (dp[i][k] + dp[i-1][(k-amari)%13]) % MOD
                else:
                    amari = (int(c) * pow(10, i, 13)) % 13
                    dp[i][k] += (dp[i][k] + dp[i - 1][(k-amari) % 13]) % MOD
 
    dprint(dp)
    print(dp[N-1][5])

solve()