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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

#define Mod 1000000007

long long div_mod(long long x, long long y, long long z)
{
	if (x % y == 0) return x / y;
	else return (div_mod((1 + x / y) * y - x, (z % y), y) * z + x) / y;
}

long long pow_mod(long long n, long long k)
{
	if (k == 0 || n == 1) return 1;
	
	long long i, N;
	for (i = 1, N = n; i * 2 <= k; i *= 2) N = N * N % Mod;
	return N * pow_mod(n, k - i) % Mod;
}

int main()
{	
	char S[100001];
	scanf("%s", S);
	
	int l, count = 0;
	for (l = 0; S[l] != 0; l++) {
		if (S[l] == '?') count++;
	}
	long long p[4] = {pow_mod(3, count), div_mod(p[0], 3, Mod), div_mod(p[1], 3, Mod), div_mod(p[2], 3, Mod)};

	int i, tmp;
	long long ans = 0, num[3] = {};
	for (i = l - 1; i >= 0; i--) {
		if (S[i] == 'C') num[2]++;
		else if (S[i] == 'B') num[1] += num[2];
		else if (S[i] == 'A') num[0] += num[1];
	}
	ans = (ans + num[0] % Mod * p[0]) % Mod;
	if (count >= 1) {
		for (i = l - 1, num[0] = 0, num[1] = 0, num[2] = 0; i >= 0; i--) {
			if (S[i] == '?') num[2]++;
			else if (S[i] == 'B') num[1] += num[2];
			else if (S[i] == 'A') num[0] += num[1];
		}
		for (i = l - 1, num[1] = 0, num[2] = 0; i >= 0; i--) {
			if (S[i] == 'C') num[2]++;
			else if (S[i] == '?') num[1] += num[2];
			else if (S[i] == 'A') num[0] += num[1];
		}
		for (i = l - 1, num[1] = 0, num[2] = 0; i >= 0; i--) {
			if (S[i] == 'C') num[2]++;
			else if (S[i] == 'B') num[1] += num[2];
			else if (S[i] == '?') num[0] += num[1];
		}
		ans = (ans + num[0] % Mod * p[1]) % Mod;
	}
	if (count >= 2) {
		for (i = l - 1, tmp = 0, num[0] = 0; i >= 0; i--) {
			if (S[i] == '?') tmp++;
			else if (S[i] == 'A' && tmp >= 2) num[0] += (long long)tmp * (tmp - 1) / 2;
			else if (S[i] == 'B' && tmp >= 1 && count - tmp >= 1) num[0] += (long long)tmp * (count - tmp);
			else if (S[i] == 'C' && count - tmp >= 2) num[0] += (long long)(count - tmp) * (count - tmp - 1) / 2;
		}
		ans = (ans + num[0] % Mod * p[2]) % Mod;
	}
	if (count >= 3) ans = (ans + ((long long)count * (count - 1) * (count - 2) / 6 % Mod) * p[3]) % Mod;
	
	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}