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.

Examples

Input

A??C


Output

8


Input

ABCBC


Output

3


Input

????C?????B??????A???????


Output

979596887
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;
		int[] sumA = new int[n + 1];
		int[] sumC = new int[n + 1];
		int[] sumX = new int[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[] three = new long[n + 1];
		three[0] = 1;
		for(int i = 0; i < n; i++) three[i + 1] = (three[i] * 3) % MOD;
		long ans = 0;
		for(int i = 1; i < n; i++) {
			if(s.charAt(i) == 'B' || s.charAt(i) == '?') {
				int tailC = sumC[n] - sumC[i + 1];
				int tailX = sumX[n] - sumX[i + 1];
				long head = (sumA[i] * three[sumX[i]] + sumX[i] * (sumX[i] == 0 ? 1 : three[sumX[i] - 1])) % MOD;
				long tail = (tailC * three[tailX] + tailX * (tailX == 0 ? 1 : three[tailX - 1])) % MOD;
				ans = (ans + (head * tail) % MOD) % MOD;
			}
		}
		System.out.println(ans);
	}
}
