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 Go (1.6) (judge verdict: Accepted)

// Package main provides
//
// File:  d.go
// Author: ymiyamoto
//
// Created on Sat Dec  8 20:05:49 2018
//
package main

import "fmt"

const MOD = 1e9 + 7

func main() {
	var S string
	fmt.Scan(&S)
	dp := make([][]int, len(S)+1)
	for i := range dp {
		dp[i] = make([]int, 4)
	}

	dp[0][0] = 1
	for i := 0; i < len(S); i++ {
		switch S[i] {
		case 'A':
			dp[i+1][0] = dp[i][0]
			dp[i+1][1] = (dp[i][1] + dp[i][0]) % MOD
			dp[i+1][2] = dp[i][2]
			dp[i+1][3] = dp[i][3]
		case 'B':
			dp[i+1][0] = dp[i][0]
			dp[i+1][1] = dp[i][1]
			dp[i+1][2] = (dp[i][2] + dp[i][1]) % MOD
			dp[i+1][3] = dp[i][3]
		case 'C':
			dp[i+1][0] = dp[i][0]
			dp[i+1][1] = dp[i][1]
			dp[i+1][2] = dp[i][2]
			dp[i+1][3] = (dp[i][3] + dp[i][2]) % MOD
		case '?':
			dp[i+1][0] = 3 * dp[i][0] % MOD
			dp[i+1][1] = (3*dp[i][1] + dp[i][0]) % MOD
			dp[i+1][2] = (3*dp[i][2] + dp[i][1]) % MOD
			dp[i+1][3] = (3*dp[i][3] + dp[i][2]) % MOD
		}
	}

	fmt.Println(dp[len(S)][3])
}