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: Wrong Answer)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
 
modulo = 10 ** 9 + 7
 
S = input()
 
def nleft(S, c):
    """各位置について，それより左の文字列に含まれる文字 c の総数.
       左に'?'が含まれるときは，それらに文字を入れて得られるすべての
       文字列に渡って，文字 c の個数を合計する. """
 
    # c の個数
    n = [0 for c in S]
    for i in range(len(S)):
        n[i] = n[i-1] + (1 if S[i-1] == c else 0)

    q = 0      # '?'の個数
    pow3q = 1  # 3 ** q
    for i in range(len(n)):
        if q > 0:
            n[i] *= pow3q         # ?の埋め方の数だけ増やす
            n[i] += pow3q1 * q    # ?に入る'C'の個数
            n[i] %= modulo
        if S[i] == '?':
            q += 1
            pow3q1 = pow3q
            pow3q  = pow3q * 3 % modulo
    return n
 
def nright(S, c):
    # c の個数
    n = [0 for c in S]
    for i in range(len(S) - 1, 0, -1):
        n[i-1] = n[i] + (1 if S[i] == c else 0)

    q = 0      # '?'の個数
    pow3q = 1  # 3 ** q
    for i in range(len(n) - 1, -1, -1):
        if q > 0:
            n[i] *= pow3q         # ?の埋め方の数だけ増やす
            n[i] += pow3q1 * q    # ?に入る'C'の個数
            n[i] %= modulo
        if S[i] == '?':
            q += 1
            pow3q1 = pow3q
            pow3q  = pow3q * 3 % modulo
    return n
 
 
left  = nleft (S, 'A')
right = nright(S, 'C')
 
# 各文字が'B'のときのABC数を計算
abc = 0
for i in range(0, len(S)):
    if S[i] in 'B?':
        abc += left[i] * right[i]
 
abc = int(abc) % modulo
print(abc)