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;

static MOD: u64 = 1_000_000_007;

fn input<T:FromStr>() -> T {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().parse().ok().unwrap()
}

fn binary_power(x: u64, n: u64) -> u64 {
    if n == 0 {
      return 1;
    }
    if n % 2 == 0 {
      return (binary_power(x, n / 2) % 13).pow(2) % 13;
    } else {
      return binary_power(x, n - 1) % 13 * x % 13;
    }
}

fn remainder(n: char, digit: u64) -> Vec<u64> {
    let mut remainders = vec![0u64; 13];
    if n == '?' {
        for i in 0..10 {
            let remainder = binary_power(10, digit) * i % 13;
            remainders[remainder as usize] += 1
        }
    } else {
        let remainder = binary_power(10, digit) * (n as u64 - 48) % 13;
        remainders[remainder as usize] = 1;
    }    
    remainders
}

fn main() {
    let s: String = input();
    let s: Vec<char> = s.chars().collect();
    let mut dp: Vec<Vec<u64>> = vec![vec![0; 13]; s.len()];
    dp[s.len() - 1] = remainder(s[s.len() - 1], 0);
    for i in (0..(s.len() - 1)).rev() {
        for j in 0..13 {
            let remainders = remainder(s[i], (s.len() - 1 - i) as u64);
            for k in 0..13 {
                let r = (13 + j - k) % 13;
                dp[i][j] += dp[i+1][r] % MOD * remainders[k] % MOD;
            }
            
        }
    }
    // println!("{:?}", dp);
    println!("{:?}", dp[0][5] % MOD);
}