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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

S = gets.chomp.reverse!

n = 13
# dp[k] -> 13で割った余りがkのパターン数
dp = Array.new(n) { 0 }
dp[0] = 1

mod = 10**9 + 7
mul = 1

S.chars.each do |char|
  next_dp = Array.new(n) { 0 }

  if char == '?'
    10.times do |num|
      n.times do |k|
        rem = (num * mul + k) % n
        next_dp[rem] += dp[k]
        next_dp[rem] %= mod
      end
    end
  else
    num = char.to_i
    n.times do |k|
      rem = (num * mul + k) % n
      next_dp[rem] += dp[k]
      next_dp[rem] %= mod
    end
  end

  mul *= 10
  mul %= n
  dp = next_dp
end

puts dp[5]