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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#define int long long
#define p (int)(1e9 + 7)

int MOD(int a){
	a %= p;
	return a >= 0 ? a : a + p;
}

int power(int a, int N){
	if(N == 0){
		return 1;
	}
	else if(N % 2 == 0){
		return power(MOD(a * a), N / 2);
	}
	else{
		return MOD(a * power(a, N - 1));
	}
}

signed main(){
	int N, i, j;
	char *S = (char *)malloc(sizeof(char) * 100002);
	scanf("%s", &S[1]);
	for(N = 0; S[N] != '\0'; N++){}
	N--;
	int *dpA = (int *)malloc(sizeof(int) * (N + 1));
	int *dpAB = (int *)malloc(sizeof(int) * (N + 1));
	int *dpABC = (int *)malloc(sizeof(int) * (N + 1));
	dpA[0] = 0;
	dpAB[0] = 0;
	dpABC[0] = 0;
	for(i = 1, j = 0; i <= N; i++){
		if(S[i] == 'A'){
			dpA[i] = MOD(dpA[i - 1] + power(3, j));
		}
		else if(S[i] == '?'){
			dpA[i] = MOD(3 * dpA[i - 1] + power(3, j));
			j++;
		}
		else{
			dpA[i] = MOD(dpA[i - 1]);
		}
	}
	for(i = 1; i <= N; i++){
		if(S[i] == 'B'){
			dpAB[i] = MOD(dpAB[i - 1] + dpA[i - 1]);
		}
		else if(S[i] == '?'){
			dpAB[i] = MOD(3 * dpAB[i - 1] + dpA[i - 1]);
		}
		else{
			dpAB[i] = MOD(dpAB[i - 1]);
		}
	}
	for(i = 1; i <= N; i++){
		if(S[i] == 'C'){
			dpABC[i] = MOD(dpABC[i - 1] + dpAB[i - 1]);
		}
		else if(S[i] == '?'){
			dpABC[i] = MOD(3 * dpABC[i - 1] + dpAB[i - 1]);
		}
		else{
			dpABC[i] = MOD(dpABC[i - 1]);
		}
	}
/*	printf("dpA:\n");
	for(i = 0; i <= N; i++){
		printf("%lld ", dpA[i]);
	}
	printf("\n");
	printf("dpAB:\n");
	for(i = 0; i <= N; i++){
		printf("%lld ", dpAB[i]);
	}
	printf("\n");
	printf("dpABC:\n");
	for(i = 0; i <= N; i++){
		printf("%lld ", dpABC[i]);
	}
	printf("\n");
*/	printf("%lld\n", dpABC[N]);
	return 0;
}