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: Runtime Error)

# 153716888
MOD = (10**9) + 7
s = gets.chomp
l = s.size
m = 13
dma = {
 10  =>  1,
 9   => 10,
 12  =>  9,
 3   => 12,
 4   =>  3,
 1   =>  4
}
b = [1, 10, 9, 12, 3, 4][(l%6)]
hatena = '?'
before = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
l.times do |i|
  curr = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
  b = dma[b]
  if s[i] == hatena
    10.times do |k|
      m.times do |j|
        curr[j] = (curr[j] + before[(j+k*b)%m])%MOD
      end
    end
  else
    k = s[i].to_i
    m.times do |j|
      curr[j] = before[(j+k*b)%m]%MOD
    end
  end
  before = curr
end
p curr[8]