Problem:
Score : 400 points

Problem Statement

The ABC number of a string T is the number of triples of integers (i, j, k) that satisfy all of the following conditions:

1 ≤ i < j < k ≤ |T| (|T| is the length of T.)

T_i = A (T_i is the i-th character of T from the beginning.)

T_j = B

T_k = C

For example, when T = ABCBC, there are three triples of integers (i, j, k) that satisfy the conditions: (1, 2, 3), (1, 2, 5), (1, 4, 5). Thus, the ABC number of T is 3.

You are given a string S. Each character of S is A, B, C or ?.

Let Q be the number of occurrences of ? in S. We can make 3^Q strings by replacing each occurrence of ? in S with A, B or C. Find the sum of the ABC numbers of all these strings.

This sum can be extremely large, so print the sum modulo 10^9 + 7.

Constraints

3 ≤ |S| ≤ 10^5

Each character of S is A, B, C or ?.

Input

Input is given from Standard Input in the following format:

S

Output

Print the sum of the ABC numbers of all the 3^Q strings, modulo 10^9 + 7.

Sample Input 1

A??C

Sample Output 1

8

In this case, Q = 2, and we can make 3^Q = 9 strings by by replacing each occurrence of ? with A, B or C. The ABC number of each of these strings is as follows:

AAAC: 0

AABC: 2

AACC: 0

ABAC: 1

ABBC: 2

ABCC: 2

ACAC: 0

ACBC: 1

ACCC: 0

The sum of these is 0 + 2 + 0 + 1 + 2 + 2 + 0 + 1 + 0 = 8, so we print 8 modulo 10^9 + 7, that is, 8.

Sample Input 2

ABCBC

Sample Output 2

3

When Q = 0, we print the ABC number of S itself, modulo 10^9 + 7. This string is the same as the one given as an example in the problem statement, and its ABC number is 3.

Sample Input 3

????C?????B??????A???????

Sample Output 3

979596887

In this case, the sum of the ABC numbers of all the 3^Q strings is 2291979612924, and we should print this number modulo 10^9 + 7, that is, 979596887.

Solution in Python (3.4.3) (judge verdict: Runtime Error)

import numpy as np

s = input()

a = np.zeros(len(s))
q = np.zeros(len(s))
ab0 = np.zeros(len(s))
ab1 = np.zeros(len(s))
ab2 = np.zeros(len(s))
abc0 = np.zeros(len(s))
abc1 = np.zeros(len(s))
abc2 = np.zeros(len(s))
abc3 = np.zeros(len(s))

for i in range(len(s)):
    if s[i]=='A':
        a[i] = a[i-1]+1
        q[i] = q[i-1]
        ab0[i] = ab0[i-1]
        ab1[i] = ab1[i-1]
        ab2[i] = ab2[i-1]
        abc0[i] = abc0[i-1]
        abc1[i] = abc1[i-1]
        abc2[i] = abc2[i-1]
        abc3[i] = abc3[i-1]
    elif s[i]=='B':
        a[i] = a[i-1]
        q[i] = q[i-1]
        ab0[i] = ab0[i-1]+a[i-1]
        ab1[i] = ab1[i-1]+q[i-1]
        ab2[i] = ab2[i-1]
        abc0[i] = abc0[i-1]
        abc1[i] = abc1[i-1]
        abc2[i] = abc2[i-1]
        abc3[i] = abc3[i-1]
    elif s[i]== 'C':
        a[i] = a[i-1]
        q[i] = q[i-1]
        ab0[i] = ab0[i-1]
        ab1[i] = ab1[i-1]
        ab2[i] = ab2[i-1]
        abc0[i] = abc0[i-1]+ab0[i-1]
        abc1[i] = abc1[i-1]+ab1[i-1]
        abc2[i] = abc2[i-1]+ab2[i-1]
        abc3[i] = abc3[i-1]
    else:
        a[i] = a[i-1]
        q[i] = q[i-1]+1
        ab0[i] = ab0[i-1]
        ab1[i] = ab1[i-1]+a[i-1]
        ab2[i] = ab2[i-1]+q[i-1]
        abc0[i] = abc0[i-1]
        abc1[i] = abc1[i-1]+ab0[i-1]
        abc2[i] = abc2[i-1]+ab1[i-1]
        abc3[i] = abc3[i-1]+ab2[i-1]

m = q[len(s)-1]
n = abc0[len(s)-1] * (3**m%1000000007)
n %= 1000000007
n += abc1[len(s)-1] * (3**(m-1)%1000000007)
n %= 1000000007
n += abc2[len(s)-1] * (3**(m-2)%1000000007)
n %= 1000000007
n += abc3[len(s)-1] * (3**(m-3)%1000000007)
n %= 1000000007
print(int(n))