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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;



class Myon
{
    public Myon() { }
    public static int Main()
    {
        cin = new Scanner();
        new Myon().calc();
        return 0;
    }

    static Scanner cin;

    void calc()
    {
        string S = cin.next();

        const int N = 13;
        //dp[k] -> 13で割った余りがKのパターン数
        long[] dp = new long[N];
        dp[0] = 1;

        //前から解くアルゴリズム
        // 1?2?3   0 -> 1 -> 1? -> 1?2 -> ...
        long mod = 1000000007;

        for (int i = 0; i < S.Length; i++)
        {
            //次の状態
            long[] nextDP = new long[N];

            char c = S[i];
            if(c == '?')
            {
                for (int k = 0; k < 10; k++)
                {
                    for (int j = 0; j < N; j++)
                    {
                        nextDP[((j * 10) + k) % N] += dp[j];
                        nextDP[((j * 10) + k) % N] %= mod;
                    }
                }
            }
            else
            {
                int k = (int)(S[i] - '0');
                for (int j = 0; j < N; j++)
                {
                    nextDP[((j * 10) + k) % N] += dp[j];
                    nextDP[((j * 10) + k) % N] %= mod;
                }
            }

            dp = nextDP;
        }

        Console.WriteLine(dp[5]);
    }
}



class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}