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)

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()
}

fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

fn main() {
    let s: String = read();
    let cv: Vec<char> = s.chars().rev().collect();
    let mut dp: Vec<Vec<u64>> = vec![vec![0; 13]; 100001];
    let l: usize = s.len();
    let rs: Vec<u64> = vec![1, 10, 9, 12, 3, 4];

    dp[0][0] = 1;
    for i in 0..l {
        let c = cv[i];
        let r = rs[i % 6];

        if c == '?' {
            for n in 0..10 {
                let dif = ((n * r) % 13) as usize;
                for j in 0..13 {
                    dp[i + 1][(j + dif) % 13] += dp[i][j];
                }
            }
        } else {
            let n = (c as u64) - 48;
            let dif = ((n * r) % 13) as usize;
            for j in 0..13 {
                dp[i + 1][(j + dif) % 13] += dp[i][j];
            }
        }
        for j in 0..13 {
            dp[i + 1][j] %= 1000000007;
        }
    }
    println!("{}", dp[l][5]);
}