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)

def main():
    s = input()
    pa = [0]
    pb = [0]
    pc = [0]
    pd = [0]

    n = len(s)
    for i in range(n):
        pa.append(pa[-1])
        pb.append(pb[-1])
        pc.append(pc[-1])
        pd.append(pd[-1])

        if s[i] == 'A':
            pa[-1] += 1
        elif s[i] == 'B':
            pb[-1] += 1
        elif s[i] == 'C':
            pc[-1] += 1
        elif s[i] == '?':
            pd[-1] += 1

    totq = pd[n]

    MOD = 1000000007
    ans = 0
    for i in range(1, n + 1):
        if s[i - 1] == 'A' or s[i - 1] == 'C':
            continue

        if s[i - 1] == 'B':
            buf = 0
        else:
            buf = 1

        # ABC ?BC AB? ?B?
        a_bef = pa[i - 1]
        d_bef = pd[i - 1]
        c_aft = pc[n] - pc[i]
        d_aft = pd[n] - pd[i]

        if a_bef * c_aft > 0:
            ans += (a_bef * c_aft % MOD) * pow(3, totq - buf, MOD) % MOD
            ans %= MOD

        if d_bef * c_aft > 0:
            ans += (d_bef * c_aft % MOD) * pow(3, totq - buf - 1, MOD) % MOD
            ans %= MOD

        if a_bef * d_aft > 0:
            ans += (a_bef * d_aft % MOD) * pow(3, totq - buf - 1, MOD) % MOD
            ans %= MOD
        
        if d_bef * d_aft > 0:
            ans += (d_bef * d_aft % MOD) * pow(3, totq - buf - 2, MOD) % MOD
            ans %= MOD

    print(ans)

main()