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::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: expr) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}
const MOD: usize = 1000000007;

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    let mut it = s.split_whitespace();
    let s = it.next().unwrap();
    let mut dp = vec![vec![0; 13]; s.len() + 1];
    dp[0][0] = 1;
    for (i, c) in s.chars().enumerate() {
        if let Some(x) = c.to_digit(10) {
            let x = x as usize;
            for j in 0..13 {
                dp[i + 1][(10 * j + x) % 13] += dp[i][j];
                dp[i + 1][(10 * j + x) % 13] %= MOD;
            }
        } else {
            for x in 0..10 {
                for j in 0..13 {
                    dp[i + 1][(10 * j + x) % 13] += dp[i][j];
                    dp[i + 1][(10 * j + x) % 13] %= MOD;
                }
            }
        }
    }
    println!("{}", dp[s.len()][5]);
}