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)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
    let mut stdin = String::new();
    std::io::Read::read_to_string(
        &mut std::io::stdin(),
        &mut stdin).unwrap();
    let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    
    let s = get().as_bytes();
    let mut dp = vec![0; 13];
    let n = s.len();
    if s[n-1] == b'?' {
        for i in 0 .. 10 {
            dp[i] = 1;
        }
    } else {
        let x = (s[n-1] - b'0') as usize;
        dp[x] = 1;
    }
    let mut p = 10;
    let md = 1_000_000_007;
    for i in 1 .. s.len() {
        let i = (s.len()-1) - i;
        let mut tmp = vec![0; 13];
        if s[i] == b'?' {
            for c in 0 .. 10 {
                let x = ((c * p) % 13) as usize;
                for j in 0 .. 13 {
                    tmp[(j + x) % 13] += dp[j];
                    tmp[(j + x) % 13] %= md;
                }
            }
        } else {
            let c = (s[i] - b'0') as i64;
            let x = ((c * p) % 13) as usize;
            for j in 0 .. 13 {
                tmp[(j + x) % 13] = dp[j];
            }
        }
        p = (p * 10) % 13;
        dp = tmp;
    }
    let ans = dp[5];
    println!("{}", ans);
}