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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdio.h>
#include<cstring>
#include<math.h>
#include<map>
#include<bitset>
#include<iomanip>
#include<queue>
#include<functional>
#include<stack>
#include<tuple>
#include<cassert>
using namespace std;

#define rep(i, n) for (long long i = 0; i < (long long)(n); ++i)
using P = pair<long long, long long>;

const long long MOD = 1000000007;
long long dp[100001][13];
long long M[6];

int main() {
  M[0] = 1; 
  for(long long i = 0; i < 5; i++) {
    M[i+1] = (M[i] * 10) % 13;
  }
  string S;
  cin >> S;
  if(S[S.size()-1] == '?') {
    for(long long i = 0; i < 13; i++) {
      dp[1][i] = 1;
    }
  } else {
    dp[1][S[S.size()-1] - '0'] = 1;
  }

  for(long long i = S.size() - 2; i >= 0; i--) {
    long long key = (S.size() - i - 1) % 6;
    if(S[i] != '?') {
      long long temp = S[i] - '0';
      for(long long j = 0; j < 13; j++) {
        long long t = (temp * M[key] + j) % 13;
        dp[S.size()-i][t] += dp[S.size()-i-1][j];
        dp[S.size()-i][t] %= MOD;
      }
    } else {
      rep(j, 10) {
        long long temp = j;
        rep(k, 13) {
          long long t = (temp * M[key] + k) % 13;
          dp[S.size()-i][t] += dp[S.size()-i-1][k];
          dp[S.size()-i][t] %= MOD;
        }
      }
    }
  }

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

  return 0;
}