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: Accepted)

"""
We Love ABC
DP使う問題
T : 文字列
i < j < k < len(T)
Ti = "A"
Tj = "B"
Tk = "C"
i j k の順で取れたら、1カウント
- TはA, B, C, ? で構成される
- ? は　ABC のいずれか
これらを満たす場合の数

i番目の文字において、ABCのどこまで進んだか
null:   0
A   :   1
B   :   2
C   :   3

?の扱いが厄介
?があると単純に選択肢が3倍になる
A? -> AA, AB, AC
B? -> BA, BB, BC
まだ選んでいない時 : 入力の選択肢が3倍に増える
dp[i+1][1]: 
    すでにAを選んでいる時 : iまでのAの選び方が3倍に増えるので、dp[i][1]*3
    何も選んでいない場合  : ?がAになれるので、dp[i][0]を加える
ほかも同じ
"""
mod = 10**9 + 7

S = input()
N = len(S)
dp = [[0]*4 for i in range(100800)]
dp[0][0] = 1
for i in range(N):
    s = S[i]
    dp[i+1][0] = dp[i][0] 
    dp[i+1][1] = dp[i][1] 
    dp[i+1][2] = dp[i][2] 
    dp[i+1][3] = dp[i][3] 
    if s == "A":
        # Aを選ぶ選択肢がそれまでのA,?分増える
        dp[i+1][1] += dp[i][0]
    if s == "B":
        # Bを選ぶ選択肢がそれまでのAの選択肢分増える
        dp[i+1][2] = dp[i][2] + dp[i][1] 
    if s == "C":
        # Cを選ぶ選択肢がBまでを選んだ分だけ増える
        dp[i+1][3] = dp[i][3] + dp[i][2]
    if s == "?":
        # ?を選んだ場合。選択肢は3倍に広がる
        dp[i+1][0] = (3*dp[i][0])%mod
        dp[i+1][1] = (dp[i][0] + 3*dp[i][1]) % mod
        dp[i+1][2] = (dp[i][1] + 3*dp[i][2]) % mod
        dp[i+1][3] = (dp[i][2] + 3*dp[i][3]) % mod
        
# Cまで行った数を求める
print(dp[N][3]%mod)