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)

MOD = 10**9+7
s = gets.chomp.split("").map{ |x|
  if x != "?"
    x.to_i
  else
    x.to_s
  end
}
len = s.length
num = Array.new(len+1).map{Array.new(13,0)}

num[0][0] = 1
i = 0
s.each do |x|
  if x == "?"
    tmp = [0]*13
    10.times do |j|
      13.times do |k|
        tmp[(k*10+j)%13] += num[i][k] % MOD
      end
      num[i+1]=tmp
    end
    # 10.times do |j|
    #   13.times do |k|
    #     num[i+1][(k*10+j)%13] += num[i][k]
    #   end
    # end
  else
    13.times do |k|
      num[i+1][(k*10+x)%13] += num[i][k] % MOD
      # num[i+1][(k*10+x)%13] += num[i][k]
    end
  end
  # 13.times do |k|
  #   num[i+1][k] %= MOD
  # end
  i += 1
end

puts num[len][5] % MOD
# puts num[len][5]