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)

s = gets.strip.split('').reverse.map {|c| c=='?' ? nil : c.to_i}
N=13
MOD = 10**9+7

dp = Array.new(13,0)
dp[0]=1
a9=(0..9).to_a
mul = 1
s.each do |num|
  ndp = Array.new(13,0)
  if num.nil?
    a9.each do |candidate|
      N.times do |pre|
        bigger=mul*candidate
        ndp[(bigger + pre)%N] = (dp[pre] + ndp[(bigger + pre)%N])
      end
    end
  else
    N.times do |pre|
        bigger=mul*num
        ndp[(bigger + pre)%N] = (dp[pre] + ndp[(bigger + pre)%N])
      end
  end
  dp=ndp.map {|e| e%MOD }
  mul = 10 * mul % N
end

p dp[5]%MOD