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)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
 
#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
 
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
 
    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
 
#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };
 
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
 
    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };
 
    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };
 
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
 
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
fn main() {
    input! {
        s: chars,
    }
    //let s = vec!['?'; 100000];
    let d = 1e9 as u64 + 7;
    let n = 13;

    let mut dp = vec![0; n];

    dp[0]= 1;

    let mut mul = 1;
    for v in s.iter().rev() {
        let mut next_dp = vec![0; n];
        if *v == '?' {
           for i in 0..10 {
               for j in 0..n {
                   next_dp[(i*mul+j)%n] = next_dp[(i*mul+j)%n] + dp[j];
                   next_dp[(i*mul+j)%n] = next_dp[(i*mul+j)%n] % d;
                   //next_dp[(i*mul+j)%n] += dp[j];
                   //next_dp[(i*mul+j)%n] %= d;
               }
           } 
        } else {
            let i = v.to_digit(10).unwrap() as usize;

            for j in 0..n {
                next_dp[(i*mul+j)%n] = next_dp[(i*mul+j)%n] + dp[j];
                next_dp[(i*mul+j)%n] = next_dp[(i*mul+j)%n] % d;
                //next_dp[(i*mul+j)%n] += dp[j];
                //next_dp[(i*mul+j)%n] %= d;
            }

        }

        for i in 0..n {
            dp[i] = next_dp[i];
        }
        mul = mul * 10;
        mul = mul % n;
    }

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

}