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: Accepted)

MOD = 10**9 + 7
s = gets.chomp

s.reverse!

rem = [1,10,9,12,3,4]
count = Array.new(2){Array.new(13,0)}
count[1][0] = 1

(s.length).times do |i|
  if s[i] == '?'
    count[i%2] = Array.new(13,0)
    sum = count[(i-1)%2].reduce(&:+)
    13.times do |j|
      count[i%2][j] = sum - count[(i-1)%2][(j+rem[i%6])%13] - count[(i-1)%2][(j+2*rem[i%6])%13] - count[(i-1)%2][(j+3*rem[i%6])%13]
      count[i%2][j] %= MOD
    end
  else
    n = s[i].to_i
    count[i%2] = count[(i-1)%2].rotate(-(n*rem[i%6]))
  end
end

puts(count[(s.length-1)%2][5])