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 <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0; i < (int)n; ++i)
#define FOR(i, a, b) for(int i = a; i < (int)b; ++i)
#define rrep(i, n) for(int i = ((int)n - 1); i >= 0; --i)

typedef long long ll;
typedef long double ld;

const ll INF = 1e18;
const int Inf = 1e9;
const double EPS = 1e-9;
const ll MOD = 1e9 + 7;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(0);
    string s;
    cin >> s;
    reverse(s.begin(), s.end());
    int l = s.size();
    vector<vector<int> > dp(l + 1, vector<int>(13));
    int p = 13;
    ll m = 1;
    dp[0][0] = 1;
    rep (i, s.size()) {
        ll t;
        if (s[i] == '?') {
            rep (j, 10) {
                t = m * j % p;
                rep (k, 13) {
                    dp[i + 1][(k + t) % p] += dp[i][k];
                    dp[i + 1][(k + t) % p] %= MOD;
                }
            }
        } else {
            t = m * (s[i] - '0') % p;
            rep (j, 13) {
                dp[i + 1][(j + t) % p] += dp[i][j];
                dp[i + 1][(j + t) % p] %= MOD;
            }
        }
        m *= 10;
        m %= p;
    }
    cout << dp[l][5] << endl;
    
    return 0;
}