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)

/*
title:

url:
*/

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <functional>
#include <queue>
#include <limits>
#include <string>
#include <utility>
#include <cstring>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); i++)
#define INF 1000000007

typedef long long ll;
typedef pair<int, int> pint;

//動的計画法を使うといいらしい。
//ちゃんとその発想を思いつけるようにしたい...
//計算量が多いがパラメーター自体は少ない、そして左から順番に読んでいく、
//という点が動的計画法の発想のポイントか...

int main()
{
    string s;
    cin >> s;
    ll dp[s.size() + 1][4];
    dp[s.size()][0] = 0;
    dp[s.size()][1] = 0;
    dp[s.size()][2] = 0;
    dp[s.size()][3] = 1;

    for (int i = s.size() - 1; i >= 0; i--)
    {
        for (int j = 0; j < 3; j++)
        {
            if (s[i] == '?')
            {
                dp[i][j] = 3 * dp[i + 1][j] + dp[i + 1][j + 1];
            }
            else
            {
                dp[i][j] = dp[i + 1][j];
                if ((s[i] - 'A') == j)
                {
                    dp[i][j] += dp[i + 1][j + 1];
                }
            }
        }
        if (s[i] == '?')
        {
            dp[i][3] = 3 * dp[i + 1][3];
        }
        else
        {
            dp[i][3] = dp[i + 1][3];
        }
        for (int j = 0; j < 4; j++)
        {
            dp[i][j] %= INF;
        }
    }

    cout << dp[0][0] << endl;
}