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 <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <tuple>
#include <cmath>
#include <cstring>
#include <cstdio>

using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;

const ll MOD = 1000000007;
int A[100000];
ll dp[100000][13];

int main() {
    string S;
    cin >> S;
    int N = S.size();
    reverse(S.begin(), S.end());
    rep(i, N) {
        if (S[i] == '?') A[i] = -1;
        else A[i] = S[i] - '0';
    }
    int D = 1;
    dp[0][0] = 1;
    for (int j = 1; j < 13; j ++) dp[0][j] = 0;
    rep(i, N) {
        if (A[i] == -1) {
            rep(j, 13) {
                ll num = 0;
                rep(x, 10) {
                    int d = x * D % 13;
                    num = (num + dp[i][(j + 13 - d) % 13]) % MOD;
                }
                dp[i + 1][j] = num;
            }
        } else {
            int d = A[i] * D % 13;
            rep(j, 13) {
                dp[i + 1][j] = dp[i][(j + 13 - d) % 13];
            }
        }
        D = (D * 10) % 13;
    }
    cout << dp[N][5] << endl;


    return 0;
}