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::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let s: Vec<char> = input::<String>().chars().collect();
    const MOD: i64 = 1000000007;

    let n = s.len();
    let mut dp = vec![vec![0i64; 13]; n + 1];
    dp[0][0] = 1;
    for i in 0..n {
        if s[i] == '?' {
            for k in 0..10 {
                for j in 0..13 {
                    let m = (j * 10 + k) % 13;
                    dp[i + 1][m] += dp[i][j];
                    dp[i + 1][m] %= MOD;
                }
            }
        } else {
            let d = (s[i] as u8 - b'0') as usize;
            for j in 0..13 {
                let m = (j * 10 + d) % 13;
                dp[i + 1][m] += dp[i][j];
                dp[i + 1][m] %= MOD;
            }
        }
    }

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