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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>
#define N 10010
#define mod 1000000007
int main(void)
{
	char S[N];
	int s[N+1];
	scanf("%s",S);
	int i,j,f,w,b;
	i = 0;
	while((int)S[i] != 0)
	{
		s[i+1] = (int)S[i] - 48;
		if(s[i+1] == 15) s[i+1] = 100;
		i++;
	}
	int keta = i;
	long long int dp[keta+1][13];
	for(i = 1; i < keta+1;i++)
	{
		for(j = 0; j < 13; j++)
		{
			dp[i][j] = 0;
		}
	}
	for(i = 0; i < 10; i++)
	{
		if(s[1] == i) dp[1][i] = 1;
		else if(s[1] == 100) dp[1][i] = 1;
	}
	for(i=2; i < keta+1; i++)
	{
		for(j=0;j<13;j++)
		{
			for(f = 0; f < dp[i-1][j];f++)
			{
				if(s[i]<20){
					w = (j*10+s[i])%13;
					dp[i][w]++;
				}
				else{
					for(b=0;b<10;b++)
					{
						w = (j*10+b)%13;
						dp[i][w]++;
						if(dp[i][w]==mod) dp[i][w] = 0;
					}
				}
			}
		}
	}
	printf("%d",dp[keta][5]);
	/*
	printf("keta = %d\n",keta);
	for(i = 1; i < keta+1;i++)
	{
		for(j = 0; j < 13; j++)
		{
			printf("dp[%d][%d] = %d\n",i,j,dp[i][j]);
		}
	}
	for(i = 1; i < keta+1; i++)
	{
		printf("s[%d] = %d\n",i,s[i]);
	}
	*/
	/*
	i = 0;
	while(i < keta)
	{
		printf("s[%d] = %d\n",i,s[i]);
		i++;
	}
	printf("keta = %d\n",keta);
	*/
	return 0;
}