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>
#define rep(i,n)for(ll i=0;i<n;i++)
#define ll long long
#define ggr getchar();getchar();return 0;
#define prique priority_queue
#define inf 1000000007
#define double long double
using namespace std;
typedef pair<ll, ll>P;
bool prime(int n) {
	int cnt = 0;
	for (int i = 1; i <= sqrt(n); i++) {
		if (n % i == 0)cnt++;
	}
	if (cnt != 1)return false;
	else return n != 1;
}
ll gcd(ll x, ll y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
ll lcm(ll x, ll y) {
	return x * y / gcd(x, y);
}
ll mod_pow(ll x,ll n,ll mod){
	ll bit=1;
	while(n){
		if(n&1)bit=bit*x%mod;
		x=x*x%mod;
		n/=2;
	}
	return bit;
}
ll dp[114514][13] = {};
signed main() {
	string s;
	cin>>s;
	ll n=s.size();
	string t = s;
    reverse(t.begin(), t.end());
	// for(int i=n-1;i>=0;i--)t[i]=s[n-1-i];
	if(t[0]=='?'){
		rep(i,10)dp[0][i]=1;
	}
	else dp[0][(t[0]-'0')]=1;
	for(ll i=0;i<n-1;i++){
		for(ll j=0;j<13;j++){
		if(t[i+1]=='?'){
			for(ll k=0;k<=9;k++){
				dp[i+1][(j+(mod_pow(10,i+1,13)*k)%13)%13]+=dp[i][j];
				dp[i+1][(j+(mod_pow(10,i+1,13)*k)%13)%13]%=inf;
			}
		}
		else{
			dp[i+1][(j+(mod_pow(10,i+1,13)*(t[i+1]-'0'))%13)%13]+=dp[i][j];
			dp[i+1][(j+(mod_pow(10,i+1,13)*(t[i+1]-'0'))%13)%13]%=inf;
	      }
	   }
    }
	cout<<dp[n-1][5]%inf<<endl;
	ggr
}