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)

C197 = (10**9)+7

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

dpc = Array.new(13, 0)
dpn = Array.new(13, 0)
dpc[0] = 1
cmod = 1

for i in 0..input.length-1 do
  if input[i] != '?' then # 固定桁
    m = (input[i].to_i)*cmod % 13
    for n in 0..12 do
      dpn[(n+m)%13] += dpc[n]
    end
  else                    # 可変桁
    (0..9).each do |v|
      m = v * cmod % 13
      (0..12).each do |n|
        dpn[(n+m)%13] += dpc[n]
      end
    end
    (0..12).each do |n|
      dpn[n] %= (C197)
    end
  end
  cmod = (cmod*10)%13
  dpc = dpn
  dpn = Array.new(13, 0)
end

print dpc[5]