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.*;
 
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		long[] as = new long[s.length()];
		long[] cs = new long[s.length()];
		long[] hleft = new long[s.length()];
		long[] hright = new long[s.length()];
		long[] pow = new long[s.length()+1];
		long MOD = 1000000007;
		as[0] = 0;
		for(int i=1;i<s.length();++i){
			as[i] += as[i-1];
			hleft[i] += hleft[i-1];
			char a = s.charAt(i-1);
			if(a=='A') as[i]++;
			if(a=='?') hleft[i]++;
		}
		cs[s.length()-1] = 0;
		for(int i=s.length()-2;i>=0;--i){
			cs[i] += cs[i+1];
			hright[i] += hright[i+1];
			char c = s.charAt(i+1);
			if(c=='C') cs[i]++;
			if(c=='?') hright[i]++;
		}
		pow[0] = 1;
		for(int i=1;i<s.length();++i){
			pow[i] = pow[i-1]*3;
			pow[i] %= MOD;
		}

		long ans = 0;
		for(int i=1;i<s.length()-1;++i){
			char b = s.charAt(i);
			if(b=='B'||b=='?'){
				long left = as[i] * pow[(int)hleft[i]];
				if(hleft[i]!=0) left += hleft[i] * pow[(int)hleft[i]-1];
				left %= MOD;
				long right = cs[i] * pow[(int)hright[i]];
				if(hright[i]!=0) right += hright[i] * pow[(int)hright[i]-1];
				right %= MOD;
				ans += left * right;
				ans %= MOD;
			}
		}

		System.out.println(ans);
		return;
	}
}