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 (GCC 5.4.1) (judge verdict: Runtime Error)


#include <stdio.h>
#include <stdlib.h>
#include <string.h>



int main(void)
{
    char s[3000];
    scanf("%s", s);
    int n;
    n = strlen(s);
    
    int i,j,k,plus,z,last;
    long long ans[2][13];
    int amari[100000];
    int now=1;
    for(i=0; i<100000; i++){
        amari[i] = now;
        now = (now * 10) % 13;
    }
    
    char ch;
    ans[0][0] = 1;
    
    for(i=0; i<n; i++){
        ch = s[n-1-i];
        for(j=0; j<13; j++){
            ans[(i+1)%2][j] = 0;
        }
        if(ch=='?'){
            for(k=0; k<10; k++){
                plus = (amari[i] * k)%13;
                for(j=0; j<13; j++){
                    ans[(i+1)%2][(j+plus)%13] += ans[i%2][j];
                }
            }
            for(j=0; j<13; j++){
                ans[(i+1)%2][(j+plus)%13] %= 1000000007;
            }
        }else{
            z = (int)ch-(int)'0';
            plus = (amari[i] * z)%13;
            for(j=0; j<13; j++){
                ans[(i+1)%2][(j+plus)%13] = ans[i%2][j];
            }
        }
        last = (i+1)%2;
    }
    
        
        
    printf("a:%lld",ans[last][5]);
}