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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>

using namespace std;

typedef long long ll;

#define MOD 1000000007

int main() {
    string s;   cin >> s;
    ll n;   n = s.length();
    ll **dp = new ll *[n+1];
    for (ll i = 0; i <= n; i++) {
        dp[i] = new ll[13];
        for (ll j = 0; j < 13; j++)
            dp[i][j] = 0;
    }

    if (s[0] == '?') {
        for (ll i = 0; i < 10; i++)
            dp[1][i] = 1;
    }
    else {
        ll y = s[0] - '0';
        dp[1][y] = 1;
    }
    
    for (ll i = 1; i < n; i++) {
        if (s[i] == '?') {
            for (ll y = 0; y < 10; y++) {
                for (ll r = 0; r < 13; r++) {
                    dp[i+1][(10*r+y)%13] += dp[i][r];
                    dp[i+1][(10*r+y)%13] %= MOD;
                }
            }
        }
        else {
            ll y = s[i] - '0';
            for (ll r = 0; r < 13; r++) {
                dp[i+1][(10*r+y)%13] += dp[i][r];
                dp[i+1][(10*r+y)%13] %= MOD;
            }
        }
    }

    cout << dp[n][5] << endl;

    return 0;
}