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
def pow(x,n)
  if n == 0
    1
  elsif n % 2 == 0
    pow(x*x%13, n/2) % 13
  else
    x * pow(x, n-1) % 13
  end
end
s = gets.chomp
len = s.length
num = Array.new(len+1,0).map{Array.new(13,0)}

if s[len-1] == "?"
  10.times do |j|
    num[1][j % 13] =  1
  end
else
  num[1][s[len-1].to_i % 13] = 1
end

for i in 1..len-1
  if s[len-1-i] == "?"
    10.times do |j|
      index = pow(10,i) * j % 13
      13.times do |k|
        num[i+1][(k+index)%13] += num[i][k]
        num[i+1][(k+index)%13] %= MOD
      end
    end
  else
    index = pow(10,i) * s[len-1-i].to_i % 13
    13.times do |k|
      num[i+1][(k+index)%13] = num[i][k]
      num[i+1][(k+index)%13] %= MOD
    end
  end
end

puts num[len][5]