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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		int n = s.length();
		final long MOD = 1000000007L;
		long[] sumA = new long[n + 1];
		long[] sumC = new long[n + 1];
		long[] sumX = new long[n + 1];
		for(int i = 0; i < n; i++) {
			if(s.charAt(i) == 'A') sumA[i + 1] = sumA[i] + 1;
			else sumA[i + 1] = sumA[i];
			if(s.charAt(i) == 'C') sumC[i + 1] = sumC[i] + 1;
			else sumC[i + 1] = sumC[i];
			if(s.charAt(i) == '?') sumX[i + 1] = sumX[i] + 1;
			else sumX[i + 1] = sumX[i];
		}
		long ans = 0;
		for(int i = 1; i < n; i++) {
			if(s.charAt(i) == 'B' || s.charAt(i) == '?') {
				long tailC = sumC[n] - sumC[i + 1];
				long tailX = sumX[n] - sumX[i + 1];
				long head = (sumA[i] * (long)Math.pow(3, sumX[i]) + sumX[i] * (long)Math.pow(3, sumX[i] - 1)) % MOD;
				long tail = (tailC * (long)Math.pow(3, tailX) + tailX * (long)Math.pow(3, tailX - 1)) % MOD;
				ans = (ans + (head * tail) % MOD) % MOD;
			}
		}
		System.out.println(ans);
	}
}