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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

const unsigned long long MOD = 1000000007LL;
long long GCD(long long a, long long b){return b == 0 ? a : GCD(b, a % b);}
long long fast_exp(long long base, long long exp) {
    long long tot = 1;
    for(;exp > 0;exp >>= 1) {
       if((exp & 1) == 1) tot = tot * base % MOD;
       base = base * base % MOD;
    }
    return tot;
}

int main(){

    cin.sync_with_stdio(0); cin.tie(0); cout.tie(0);

    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    string s;
    cin >> s;

    long long tot = 0, a = 0, b = 0, c = 0, q = 1, INV = fast_exp(3, MOD - 2) % MOD;

    for(auto i : s){
        if(i == 'A'){
            ++a;
        } else if(i == 'B'){
            b += a;
        } else if(i == 'C'){
            c += b;
        } else {
            c += b * INV;
            b += a * INV;
            a += INV;
            q *= 3;
        }
        a %= MOD; b %= MOD; c %= MOD; q %= MOD;
    }

    cout << c * q % MOD << "\n";

    return 0;

}