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 Python (3.4.3) (judge verdict: Runtime Error)

S = str(input())
mod = 10**9 + 7
odd = ""
oddCnt = [0,0,0]
oddSum = 0
even = ""
evenCnt = [0,0,0]
evenSum = 0
for i in range(len(S)//3):
    for j in reversed(range(3)):
        if S[i+j] == "?":
            if i%2 == 0:
                odd += "0"
                oddCnt[j] += 1
            else:
                even += "0"
                evenCnt[j] += 1
        else:
            if i%2 == 0:
                odd += S[i+j]
            else:
                even += S[i+j]
    if odd != "":
        oddSum += int(odd)
    if even != "":
        evenSum += int(even)
    odd = ""
    even = ""
if len(S)%3 != 0 and (len(S)//3 + 1)%2 == 1:
    for j in reversed(range(len(s)%3)):
        if S[(len(s)//3) * 3+j] == "?":
            odd += "0"
            oddCnt[j] += 1
        else:
            odd += S[(len(s)//3) * 3+j]
    oddSum += int(odd)
elif len(S)%3 != 0 and (len(S)//3 + 1)%2 == 0:
    for j in reversed(range(len(s)%3)):
        if S[(len(s)//3) * 3+j] == "?":
            even += "0"
            evenCnt += 1
        else:
            even += S[(len(s)//3) * 3+j]
    evenSum += int(even)
tmpL = oddSum - evenSum + (oddCnt[2]+evenCnt[2])*900 + (oddCnt[1]+evenCnt[1])*90 + (oddCnt[0]+evenCnt[0])*9
tmpS = oddSum - evenSum - (oddCnt[2]+evenCnt[2])*900 + (oddCnt[1]+evenCnt[1])*90 + (oddCnt[0]+evenCnt[0])*9
print(((tmpL-5)//13 - (tmpS-6)//13)%mod)