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 <stdio.h>
#include <iostream>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>

using namespace std;

#define rep(i,x) for(int i=0;i<x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define FIND(x, target) find(x.begin(),x.end(),target)
#define pb(x) push_back(x)
#define output(x) cout<<x<<endl
#define ERASE(x, n) x.erase(x.begin(), x.begin()+n)
// #define int long long
#define P pair<int, int>
typedef long long ll;

const long long int MOD = 1e9+7;
const int INF = -1;
ll dp[100005][13];

signed main(void) {
  	string S;
  	cin >> S;

  	ll beki = 1;
  	for (int i=0; i<S.size(); i++) {
  		if (S[i] != '?') {
  			ll number = int (S[i]) - int('0');
  			if (i==0) {
  				dp[i][number%13] = 1;
  			} else {
	  			for (int j=0; j<13; j++) {
	  				dp[i][(number + j*10)%13] = dp[i-1][j]%MOD;
	  				// cout << number + j*10 << endl;
	  				// cout << i << " " << (number + j*10)%13 << ": " << dp[i][(number + j*10)%13] << endl;
	  			}
	  		}
  		} else {
  			if (i==0) {
  				for (int j=0; j<10; j++) {
			  		dp[i][j] = 1;
			  	}
  			} else {
	  			for (int j=0; j<10; j++) {
	  				for (int k=0; k<13; k++) {
	  					dp[i][(j+k*10)%13] = (dp[i][(j+k*10)%13] + dp[i-1][k])%MOD;
	  				}
	  			}
	  		}
  		}

  // 		for (int j=0; j<13; j++) {
		// 	cout << i << " " << j << ": " << dp[i][j] << endl;
		// }
  	}

  	cout << dp[S.size()-1][5] << endl;

	return 0;
}