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: Wrong Answer)

#include <cstdio>
#include <string>
#include <iostream>
using namespace std;
int main(){
	string s;
	cin >> s;
	int n = s.size();
	int a[n];
	for (int i = 0; i < n; i++){
		if (s[i] == 'A'){
			a[i] = 0;
		}else if (s[i] == 'B'){
			a[i] = 1;
		}else if (s[i] == 'C'){
			a[i] = 2;
		}else{
			a[i] = 3;
		}
	}
	int before[n][4][4], after[n][4][4];	//how many [j]s before or after [i] in kth position
	for (int i = 0; i < n; i++){
		for (int j = 0; j < 4; j++){
			for (int k = 0; k < 4; k++){
				before[i][j][k] = 0;
				after[i][j][k] = 0;
			}
		}
	}
	for (int i = 0; i < 4; i++){
		for (int j = 0; j < 4; j++){
			int count = 0;
			for (int k = 0; k < n; k++){
				if (a[k] == i){
					before[k][i][j] = count;
				}else if (a[k] == j){
					count++;
				}
			}
			count = 0;
			for (int k = n - 1; k >= 0; k--){
				if (a[k] == i){
					after[k][i][j] = count;
				}else if (a[k] == j){
					count++;
				}
			}
		}
	}
	long long int ans = 0;
	//count ABC, ?BC, AB?, ?B?
	for (int i = 0; i < n; i++){
		if (a[i] == 1){
			ans += (before[i][1][0] + before[i][1][3]) * (after[i][1][2] + after[i][1][3]);
			ans %= 1000000007;
		}
	}
	//count ??C
	for (int i = 0; i < n; i++){
		if (a[i] == 2){
			ans += before[i][2][3] * (before[i][2][3] - 1) / 2;
			ans %= 1000000007;
		}
	}
	//count A??
	for (int i = 0; i < n; i++){
		if (a[i] == 0){
			ans += after[i][0][3] * (after[i][0][3] - 1) / 2;
			ans %= 1000000007;
		}
	}
	//count A?C
	for (int i = 0; i < n; i++){
		if (a[i] == 3){
			ans += before[i][3][0] * after[i][3][2];
			ans %= 1000000007;
		}
	}
	//count ???
	int num = 0;
	for (int i = 0; i < n; i++){
		if (a[i] == 3){
			num++;
		}
	}
	ans += num * (num - 1) * (num - 2) / 6;
	ans %= 1000000007;
	printf("%lld\n",ans);
}