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: Accepted)

#include <stdio.h>
#include <unistd.h>


const int mod = 1000000007;

char s[100001];

int dp[2][13];

int main(){
  int i, k, r, b, d, l, t, sum, nsum;
 
  for(t=0;(k=read(STDIN_FILENO, s+t, sizeof(s)-t)) > 0;t+=k);

  for(l = 0; s[l] >= '0'; l++) ;

  r = 5;
  dp[0][0] = 1;
  b = 0;
  d = 1;
  sum = 1;
  for(i = l - 1; i >= 0; i--, d = d * 10 % 13){
    if(s[i] != '?'){
      r -= (s[i] - '0') * d;
      continue;
    }
    nsum = 0;
    for(k = 0; k < 13; k++){
      int a = k + d;
      if(a >= 13) a -= 13;
      dp[b^1][k] = sum;
      dp[b^1][k] -= dp[b][a];
      if(dp[b^1][k] < 0) dp[b^1][k] += mod;
      a += d;
      if(a >= 13) a -= 13;
      dp[b^1][k] -= dp[b][a];
      if(dp[b^1][k] < 0) dp[b^1][k] += mod;
      a += d;
      if(a >= 13) a -= 13;
      dp[b^1][k] -= dp[b][a];
      if(dp[b^1][k] < 0) dp[b^1][k] += mod;
      nsum += dp[b^1][k];
      if(nsum >= mod) nsum -= mod;
    }
    sum = nsum;
    b ^= 1;
  }

  r %= 13;
  if(r < 0) r += 13;

  printf("%d\n", dp[b][r]);
  
  return 0;
}