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)

input = gets.chomp.split('').reverse

dp = Array.new(input.length+1){|| Array.new(13, 0)}
dp[0][0] = 1
c = 1
for i in 0..input.length-1 do
  if input[i] != '?' then # 固定桁
    # その桁の剰余
    m = (input[i].to_i)*c % 13
    for n in 0..12 do
      # dp[i]をdp[i+1]にシフト
      dp[i+1][(n+m)%13] += dp[i][n]
    end
  else                    # 可変桁
    for v in 0..9 do
      m = v * c % 13
      for n in 0..12 do
        # dp[i]をdp[i+1]にシフト
        dp[i+1][(n+m)%13] += dp[i][n]
      end
    end
    for n in 0..12 do
      dp[i+1][n] %= ((10**9)+7)
    end
  end
  c *= 10
end

#p dp
print dp[input.length][5]