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)

# 桁DP

MOD = 10 ** 9 + 7
s = gets.chomp.split("")
length = s.length
# dp[上から i 桁まで][13 で割ったあまり]の出現回数
# 桁の自由度(smaller)は本問では不要
dp = (length + 1).times.map { Array.new(13, 0) }
# 初期化
# dp[0 桁目まで][13 で割ったあまりが 0]の出現回数は 1
dp[0][0] = 1
# （右から見た）桁数 % 6
mod_memo = [4, 1, 10, 9, 12, 3]
all_nums = (0..9).to_a

# 桁数分ループする（i は遷移元の桁）
length.times do |i|
  # 13 で割ったあまりごとに確認する
  13.times do |mod|
    # 桁の数字が "?" の場合は 10 回ループ、それ以外の場合はその数字のみ
    nums = s[i] == "?" ? all_nums : [s[i].to_i]
    # 13 で割ったあまりの増加分を求める
    addition = mod_memo[(length - i) % 6]
    original_count = dp[i][mod] % MOD
    nums.each do |d|
      next_mod = (mod + addition * d) % 13
      # 次の桁に遷移する
      dp[i + 1][next_mod] += original_count
    end
  end
end

puts dp[length][5] %= MOD