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;
use std::fmt::Debug;
use std::cmp;

#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn main(){
    let s = read_as_string();

    let mut dp: Vec<Vec<usize>> = vec![vec![0;13];s.len()+1];
    dp[0][0] = 1;
    for (i, ch) in s.bytes().enumerate(){
        let i = i+1;
        //if number
        if 48 <= ch && ch < 58{
            let n = (ch - 48) as usize;
            for j in 0..13{
                dp[i][(j*10+n)%13] += dp[i-1][j];
            }
        }else{
            for j in 0..13{
                for k in 0..10{
                    dp[i][(j*10+k)%13] += dp[i-1][j];
                }
            }
        }

        for j in 0..13{
            dp[i][j] %= 1000000007;
        }
    }

    println!("{}", dp[s.len()][5]);
}