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: Runtime Error)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
  let stdin = stdin();
  let stdin = stdin.lock();
  let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed to read char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
  token.parse().ok().expect("failed to parse token")
}

fn read_n<T: FromStr>(n: usize) -> Vec<T> {
  read_n_logic::<T>(n, vec![])
}

fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
  match n {
    0 => a,
    _ => {
      a.push(read::<T>());
      read_n_logic(n - 1, a)
    }
  }
}

fn main() {
  let n = read::<usize>();
  let mut a = read_n::<usize>(n + 1);
  let mut b = read_n::<usize>(n);
  let mut answer = 0;

  a.reverse();
  b.reverse();

  for i in 0..b.len() {
    for add in 0..2 {
      if b[i] > a[i + add] {
        answer += a[i + add];
        b[i] -= a[i + add];
        a[i + add] = 0;
      } else {
        answer += b[i];
        a[i + add] -= b[i];
        b[i] = 0;
      }
    }
  }

  println!("{}", answer);
}