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;

class Main{
	static int G;
	static int D;
	static int[] p;
	static int[] c;
	static boolean[] used;
	static long ans = 1000000000;


	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		String ss = sc.next();
		long[][] dp = new long[ss.length()+1][4];
		dp[0][0] = 1;
		long mod = 1000000007;
		for(int i = 0;i < ss.length();i++){
			if(ss.charAt(i) == 'A' ){
				dp[i+1][0] += dp[i][0] %mod;			//Aが入っている
				dp[i+1][1] += dp[i][1] % mod;			//Aが入っている
				dp[i+1][2] += dp[i][2] % mod;			//Aが入っている
				dp[i+1][3] += dp[i][3] % mod;			//Aが入っている
				dp[i+1][1] += 1 % mod;

			}else if(ss.charAt(i) == 'B'){
				dp[i+1][0] += dp[i][0] % mod;			//Aが入っている
				dp[i+1][1] += dp[i][1] % mod;			//Aが入っている
				dp[i+1][2] += dp[i][2] % mod;			//Aが入っている
				dp[i+1][3] += dp[i][3] % mod;			//Aが入っている
				dp[i+1][2] += dp[i][1] % mod;
			}else if(ss.charAt(i) == 'C'){
				dp[i+1][0] += dp[i][0] % mod;			//Aが入っている
				dp[i+1][1] += dp[i][1] % mod;			//Aが入っている
				dp[i+1][2] += dp[i][2] % mod;			//Aが入っている
				dp[i+1][3] += dp[i][3] % mod;			//Aが入っている
				dp[i+1][3] += dp[i][2] % mod;
			}
			if( ss.charAt(i) == '?'){
				dp[i+1][0] +=3*dp[i][0] % mod;			//Aが入っている
				dp[i+1][1] +=3*dp[i][1] % mod;			//Aが入っている
				dp[i+1][2] +=3*dp[i][2] % mod;			//Aが入っている
				dp[i+1][3] +=3*dp[i][3] % mod;	
				
				
				dp[i+1][1] += 1l % mod;
				dp[i+1][2] += dp[i][1] % mod;
				dp[i+1][3] += dp[i][2] % mod;



			}
//			System.out.println(dp[i+1][0] + " "+dp[i+1][1] + " "+dp[i+1][2] +" "+dp[i+1][3] );
		}

		System.out.println(dp[ss.length()][3] % mod);


	}


}

class Pair implements Comparable{
	int from;		//p
	int end;		//y
	int num;
	int bango;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return from - otherpair.from;
	}
}