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 (1.42.0) (judge verdict: Wrong Answer)

use proconio::{fastout, input};
use std::io::*;
use std::str::FromStr;

use std::cmp::{max, min};

const MOD: i64 = 1_000_000_007;

fn main() {
    input! {
        s: String,
    }

    let mut sc: Vec<char> = s.chars().collect();
    sc.reverse();
    let mut dp: Vec<Vec<usize>> = vec![vec![0; 13]; s.len() + 1];

    dp[0][0] = 1;
    let mut mul: i64 = 1;

    for i in 0..sc.len() {
        if sc[i] == '?' {
            for j in 0..10 {
                for k in 0..13 {
                    dp[i + 1][(mul as usize * j + k) % 13] += dp[i][k];
                    dp[i + 1][(mul as usize * j + k) % 13] %= MOD as usize;
                }
            }
        } else {
            let num = sc[i] as u32 - '0' as u32;
            for j in 0..13 as usize {
                dp[i + 1][(mul as i64 * num as i64 + j as i64) as usize % 13] += dp[i][j];
                dp[i + 1][(mul as i64 * num as i64 + j as i64) as usize % 13] %= MOD as usize;
            }
        }

        // for i in 0..13 {
        //     dp[i + 1][i] %= MOD as usize;
        // }
        mul *= 10;
        mul %= MOD;
    }
    println!("{}", dp[s.len()][5]);
}