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::*;
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 main() {
    let s: String = read();
    // 0: remain: 0
    let mut dp: Vec<[usize; 13]> = vec![[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]];
    let mut mul = 1;
    for (i, c) in s.chars().rev().enumerate() {
        let arr = [0; 13];
        dp.push(arr);
        if c == '?' {
            for n in 0..10 {
                let remain = (n * mul) % 13;
                for (j, v) in dp[i].clone().iter().enumerate() {
                    dp[i + 1][(j + remain) % 13] += *v;
                    dp[i + 1][(j + remain) % 13] %= 10_usize.pow(9) + 7;
                }
            }
        } else {
            let n = c.to_digit(10).unwrap() as usize;
            let remain = (n * mul) % 13;
            for (j, v) in dp[i].clone().iter().enumerate() {
                dp[i + 1][(j + remain) % 13] += *v;
                dp[i + 1][(j + remain) % 13] %= 10_usize.pow(9) + 7;
            }
        }
        mul *= 10;
        mul %= 13;
    }
    println!("{}", dp[dp.len() - 1][5]);
}