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)

def main():
    """
    1 <= i < j < k <= |T|
    Ti = A
    Tj = B
    Tk = C
    """
    S = input()
    ans3 = cumsum(S)

    print(ans3)

def cumsum(S):
    """
    考えたときに累積和が浮かんだ
    ABCとあるので
    i文字目までのA相当の数、つまり?も含んだ数
    これをBとCにも適用する
    Aを決め打ちで進めるとBも走査することになる
    Cは定数倍で求められるが、|S|^2: TLE で意味がない

    ここで断念していろいろ調べると
    Bを決め打ちすれば
    A: 0-(B-1)
    C: (B+1)-N
    なのでNでおｋ

    と、思ったがWAと同じ間違いをしている
    3^q パターンをきちんと考える
    """
    n = len(S)
    q = [0 for _ in range(n+1)]
    abc = [[0] * 3 for _ in range(n+1)]
    for i in range(n):
        is_q = S[i] == "?"
        q[i+1] += q[i] + int(is_q)

        # for j, c in enumerate("ABC"):
        #    add = int(is_q or S[i] == c)
        for j, c in enumerate("ABC"):
            if c == "B":
                add = int(is_q or S[i] == c)
            else:
                add = int(S[i] == c)
            abc[i+1][j] += abc[i][j] + add

    ans = 0
    MOD = 10 ** 9 + 7
    for i in range(1, n-1):
        a = abc[i][0]
        b = abc[i+1][1] - abc[i][1]
        c = abc[n][2] - abc[i+1][2]
        l = q[i]
        r = q[n] - q[i+1]

        # WA: ? を無視している
        # ans += a * b * c

        # ABC の作り方
        # AB?: ? 1つをCに割り当てる
        # ?BC: ? 1つをAに割り当てる
        # ?B?: ? 2つをACに割り当てる
        ac = a * b * c * (3 ** (l + r))
        aq = a * b * r * (3 ** (l + r - 1))
        qc = l * b * c * (3 ** (l + r - 1))
        qq = l * b * r * (3 ** (l + r - 2))
        ans += ac + aq + qc + qq
        ans %= MOD

    return ans


if __name__ == '__main__':
    main()