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)

#pragma GCC optimize("O3")

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>

# include <stdio.h>
# include <bits/stdc++.h>


#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define For(i,x,y)  for (ll i = x; i <= y; i ++) 
#define FOr(i,x,y)  for (ll i = x; i >= y; i --)
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
// ROAD to...                                                                                                                                                                                                                Red

using namespace __gnu_pbds;
using namespace std;
 
typedef
tree<pair<int,int>,null_type,less<pair<int,int>>,rb_tree_tag, tree_order_statistics_node_update> ordered_set;

inline bool isvowel (char c) {
	c = tolower(c);
    if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u') return 1;
    return 0;
}

const double eps = 0.000001;
const ld pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const ll MOD = 1e18 + 9;
const ll INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 1e5 + 123;                                          
const int M = 22;
const int pri = 997;
const int Magic = 2101;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
 
int n, m, k;
int cnt[N];
string s;
int dp[N][13];

void upd (int &a, int b) {
	a += b;
	if (a >= mod) a -= mod;
	
}

int main () {
	SpeedForce;

	cin >> s;
	n = sz(s);
	dp[0][0] = 1;
	for (int i = 0; i < n; i ++) {
		for (int j = 0; j < 13; j ++) {
		//	if (dp[i][j] > 0)
		//		cout << ' ' << i << ' ' << j << ' ' << dp[i][j] << '\n';
			if (s[i] == '?') {
				for (int k = 0; k < 10; k ++) {
					upd(dp[i + 1][(j * 10 + k) % 13], dp[i][j]);
				}
			} else {
				int k = s[i] - '0';
				upd(dp[i + 1][(j * 10 + k) % 13], dp[i][j]);
			}
		}
	}
	cout << dp[n][5];

	return Accepted;
}

// B...a