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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author mikit
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DWeLoveABC solver = new DWeLoveABC();
        solver.solve(1, in, out);
        out.close();
    }

    static class DWeLoveABC {
        private static final long MOD = (int) 1e9 + 7;
        private long[] pow;

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            String s = in.next();
            pow = new long[s.length()];
            int[] a = new int[s.length()];
            int[] c = new int[s.length()];
            int[] q = new int[s.length()];
            a[0] = s.charAt(0) == 'A' ? 1 : 0;
            c[0] = s.charAt(0) == 'C' ? 1 : 0;
            q[0] = s.charAt(0) == '?' ? 1 : 0;
            for (int i = 1; i < s.length(); i++) {
                a[i] = a[i - 1] + (s.charAt(i) == 'A' ? 1 : 0);
                c[i] = c[i - 1] + (s.charAt(i) == 'C' ? 1 : 0);
                q[i] = q[i - 1] + (s.charAt(i) == '?' ? 1 : 0);
            }
            int allC = c[s.length() - 1];
            int allQ = q[s.length() - 1];
            // System.out.println(Arrays.toString(a));
            // System.out.println(Arrays.toString(c));
            // System.out.println(Arrays.toString(q));

            long ans = 0;
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) == 'B') {
                    System.out.println("HERE B:" + i);
                    int aLeft = a[i];
                    int qLeft = q[i];
                    int cRight = allC - c[i];
                    int qRight = allQ - q[i];
                    // System.out.println(Arrays.toString(new int[]{aLeft, qLeft, cRight, qRight}));
                    ans += aLeft * cRight * pow(allQ);
                    ans += qLeft * cRight * pow(allQ - 1);
                    ans += aLeft * qRight * pow(allQ - 1);
                    ans += qLeft * qRight * pow(allQ - 2);
                    ans %= MOD;
                } else if (s.charAt(i) == '?') {
                    // System.out.println("HERE ?:"+i);
                    int aLeft = a[i];
                    int qLeft = q[i] - 1;
                    int cRight = allC - c[i];
                    int qRight = allQ - q[i];
                    // System.out.println(Arrays.toString(new int[]{aLeft, qLeft, cRight, qRight}));
                    ans += aLeft * cRight * pow(allQ - 1);
                    ans += qLeft * cRight * pow(allQ - 2);
                    ans += aLeft * qRight * pow(allQ - 2);
                    ans += qLeft * qRight * pow(allQ - 3);
                    ans %= MOD;
                }
            }
            out.println(ans);
        }

        long pow(int y) {
            if (y < 0) {
                return 0; // 1より小さいので実質0
            } else if (y == 0) {
                return 1;
            } else if (pow[y] != 0) {
                return pow[y];
            } else if (y % 2 == 0) {
                long z = pow(y / 2);
                return pow[y] = (z * z) % MOD;
            } else {
                return pow[y] = (3 % MOD) * pow(y - 1) % MOD;
            }
        }

    }
}