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 Rust (1.15.1) (judge verdict: Accepted)

use std::vec::Vec;
use std::cmp::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

const MOD: i64 = 1_000_000_000 + 7;

fn main() {
    let mut s: Vec<char> = read::<String>().chars().collect();
    let n: usize = s.len();
    s.reverse();
    let mut dp: Vec<Vec<i64>> = vec![vec![0; 13]; n+1];
    let mut tmp = 1;

    if s[0] == '?' {
      for j in 0..10 {
        dp[0][j] = 1;
      }
    } else {
      let num = (s[0].to_string()).parse::<usize>().unwrap();
      dp[0][num] = 1;
    }

    for i in 0..n-1 {
      tmp *= 10;
      tmp %= 13;
      if s[i+1] == '?' {
        for j in 0..13 {
          for k in 0..10 {
            let index = (j - k * tmp + 13 * 1000) % 13;
            dp[i+1][j] += dp[i][index];
            dp[i+1][j] %= MOD;
          }
        }
      } else {
        let num = (s[i+1].to_string()).parse::<usize>().unwrap();
        for j in 0..13 {
          let index = (j - num * tmp + 13 * 1000) % 13;
          //println!("j = {}, num * tmp = {}, index = {}", j, num * tmp, index);
          dp[i+1][j] += dp[i][index];
          dp[i+1][j] %= MOD;
        }
      }
    }




  println!("{}", dp[n-1][5])
}