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.*;

import static java.lang.System.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String S = sc.next();

        long[][] dp = new long[S.length() + 1][4];
        long prime = 1000000007;
        dp[0][0] = 1;
        for (int i = 1; i <= S.length(); i++) {
            for (int j = 0; j < 4; j++) {
                dp[i][j] = dp[i - 1][j];
                dp[i][j] %= prime;
            }
            if (S.charAt(i - 1) == 'A') {
                dp[i][1] += dp[i - 1][0];
                dp[i][1] %= prime;
            } else if (S.charAt(i - 1) == 'B') {
                dp[i][2] += dp[i - 1][1];
                dp[i][2] %= prime;
            } else if (S.charAt(i - 1) == 'C') {
                dp[i][3] += dp[i - 1][2];
                dp[i][3] %= prime;
            } else {
                for (int j = 0; j < 4; j++) {
                    dp[i][j] += 2 * dp[i - 1][j];
                    dp[i][j] %= prime;
                }
                dp[i][1] += dp[i - 1][0];
                dp[i][2] += dp[i - 1][1];
                dp[i][3] += dp[i - 1][2];
                dp[i][1] %= prime;
                dp[i][2] %= prime;
                dp[i][3] %= prime;
            }
        }

        out.println(dp[S.length()][3]);
    }
}