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 Julia (0.5.0) (judge verdict: Wrong Answer)

function main()
    """ver. 0.5.0

    ref:
        https://qiita.com/phigasui/items/2db20b36fb85e161e0ae
        https://nbviewer.jupyter.org/github/bicycle1885/Julia-Tutorial/blob/master/Julia%E3%82%AF%E3%83%83%E3%82%AF%E3%83%96%E3%83%83%E3%82%AF.ipynb
        https://nbviewer.jupyter.org/github/bicycle1885/Julia-Tutorial/blob/master/Julia%E9%AB%98%E9%80%9F%E3%83%81%E3%83%A5%E3%83%BC%E3%83%88%E3%83%AA%E3%82%A2%E3%83%AB.ipynb
    """
    const S = readline() |> chomp
    const ans = f(S)

    @assert ans == ans
    println(ans)
end

function f(S::String)
    const n = length(S)
    # Int: Int64
    const aa = zeros(Int, n)
    const cc = zeros(Int, n)
    const q = [0 for _ = 1:n]
    aa[1] = convert(Int, S[1] == 'A')
    cc[1] = convert(Int, S[1] == 'C')
    q[1]  = convert(Int, S[1] == '?')
    for i = 2:n
        aa[i] = aa[i - 1] + Int(S[i] == 'A')
        cc[i] = cc[i - 1] + Int(S[i] == 'C')
        q[i]  = q[i - 1]  + Int(S[i] == '?')
    end

    ans = 0
    const MOD = 10^9 + 7
    for i = 2:n - 1
        a = aa[i - 1]
        c = cc[n] - cc[i]
        l = q[i - 1] 
        r = q[n] - q[i]
        # 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
        if (S[i] == 'B' || S[i] == '?')
            ac = a * c * (3^(l + r))
            aq = (l + r - 1) < 0? 0: a * r * (3^(l + r - 1))
            qc = (l + r - 1) < 0? 0: l * c * (3^(l + r - 1))
            qq = (l + r - 2) < 0? 0: l * r * (3^(l + r - 2))
            # println(a, c, l, r, (ac, aq, qc, qq))
            ans += ac + aq + qc + qq
            ans = ans % MOD
        end
    end

    return ans
end

main()