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::str::FromStr;

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

// fn read_vec<T: FromStr>() -> Vec<T> {
//     let mut buf = String::new();
//     std::io::stdin().read_line(&mut buf).ok();
//     buf.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
// }

fn main() {
    let s: Vec<char> = read::<String>().chars().collect();
    let n = s.len();
    let mut dp = vec![[0; 14usize]; n+1];
    dp[0][0] = 1;
    for i in 0usize..n {
        for j in 0usize..14usize {
            if s[i] == '?' {
                for k in 0..10 {
                    let tmp: i64 = dp[i+1][(j*10+k) % 13];
                    dp[i+1][(j*10+k) % 13] = (tmp + dp[i][j]) % 1000000007;
                }
            }
            else {
                let k: usize = s[i] as usize - 48;
                let tmp: i64 = dp[i+1][(j*10+k) % 13];
                dp[i+1][(j*10+k) % 13] = (tmp + dp[i][j]) % 1000000007;
            }
        }
    }
    println!("{}", dp[n][5]);
}