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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;

pub fn main(){
    let mut s_str: String = String::new();
    io::stdin().read_line(&mut s_str).expect("!!");
    let s: Vec<char> = s_str.trim().chars().collect();
    let N = s.len();
    let mut dp: Vec<Vec<i64>> = vec![vec![0i64;4];N as usize];

    let abc = ['A', 'B', 'C'];

    for i in (0..N).rev(){
        for j in 0..4 {
            let c = s[i];

            if i == N - 1 {
                match j {
                    3 => {
                        dp[i][j] = if c == '?' {3} else {1};
                    },
                    2 => {
                        dp[i][j] = if c == '?' || c == 'C' {1} else {0};
                    }
                    _ => {
                        dp[i][j] = 0
                    },
                }
            }else{
                match j{
                    3 => {
                        dp[i][j] = if c == '?' {3} else {1} * dp[i+1][j];
                    },
                    _ => {
                        if c == abc[j] {
                            dp[i][j] = dp[i+1][j+1] + dp[i+1][j];
                        }else if c == '?' {
                            dp[i][j] = 3 * dp[i+1][j] + dp[i+1][j+1];
                        }else{
                            dp[i][j] = dp[i+1][j];
                        }
                    }
                }
            }
            dp[i][j] %= 10i64.pow(9) + 7;

        }
    }
    println!("{}", dp[0][0]);


}