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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <string>

using namespace std;
typedef long long ll;

struct mint{
    static const long long MOD = 1000000007;

    long long v;

    mint() : v(0) {}
    mint(long long x){
        x = x % MOD;
        if(x < 0){ x += MOD; }
        v = x;
    }
    mint& operator+=(mint a){
        v += a.v;
        if(v >= MOD){ v -= MOD; }
        return *this;
    }
    mint& operator*=(mint a){
        v *= a.v;
        v %= MOD;
        return *this;
    }
    mint& operator-=(mint a){
        v -= a.v;
        if(v < 0){ v += MOD; }
        return *this;
    }
    mint& operator/=(mint a){
        return (*this) *= a.inv();
    }

    mint operator*(mint a) const{
        return mint(v) *= a;
    }

    mint operator+(mint a) const {
        return mint(v) += a;
    }

    mint operator-(mint a) const {
        return mint(v) -= a;
    }

    mint operator/(mint a) const {
        return mint(v) /= a;
    }

    mint pow(long long k) const {
        mint res(1),tmp(v);
        while(k){
            if(k&1) res *= tmp;
            tmp *= tmp;
            k >>= 1;
        }
        return res;
    }
    mint inv(){return pow(MOD - 2); }

    static mint comb(long long n, int k){
        mint res(1);
        for(int i = 0; i < k; ++i){
            res *= mint(n - i);
            res /= mint(i + 1);
        }
        return res;
    }

    static mint factorial(long long n){
        mint res(1);
        for(int i = n; i > 1; --i){
            res *= mint(i);
        }
        return res;
    }

    bool operator<(const mint &a) const {
        return v < a.v;
    };
};

int main(){
    string s;
    getline(cin, s);
    vector<mint> x(13);
    x.at(0) = 1;

    ll p = 1;
    for(int i = s.size() - 1; i >= 0; --i){
        vector<mint> y(13);
        char c = s.at(i);
        if(c == '?'){
            for(int d = 0; d < 10; ++d){
                for(int j = 0; j < 13; ++j){
                    y.at((j + d * p) % 13) += x.at(j);
                }
            }
        }else{
            ll d = c - '0';
            for(int j = 0; j < 13; ++j){
                y.at((j + d * p) % 13) = x.at(j);
            }
        }
        x = y;
        p = (p * 10) % 13;
    }

    cout << x.at(5).v << endl;
    return 0;
}