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::str::FromStr;
use std::fmt::Debug;
use std::cmp;

#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn print_dp<T: Debug>(dp : &Vec<Vec<T>>){
    for row in dp{
        println!("{:?}", row);
    }
}

/*

AA A
   B
   C

AB A
   B
   C

AC A
   B
   C
*/

fn main(){
    let s = read_as_string();

    let mut dp: Vec<Vec<u64>> = vec![vec![0;3];s.len()+1];
    let m: u64 = 1_000_000_007;
    let mut pow: u64 = 1;
    for (i, ch) in s.chars().enumerate(){
        let i = i+1;
        let is_wild = ch == '?';
        if ch == 'A'  || is_wild {
            dp[i][0] += dp[i-1][0] + pow;
            dp[i][1] += dp[i-1][1];
            dp[i][2] += dp[i-1][2];
        }

        if ch == 'B' || is_wild{
            dp[i][0] += dp[i-1][0];
            dp[i][1] += dp[i-1][1] + dp[i-1][0];
            dp[i][2] += dp[i-1][2];
        }

        if ch == 'C' || is_wild{
            dp[i][0] += dp[i-1][0];
            dp[i][1] += dp[i-1][1];
            dp[i][2] += dp[i-1][2] + dp[i-1][1];
        }

        for j in 0..3{
            dp[i][j] %= m;
        }
        if is_wild{
            pow *= 3;
            pow %= m;
        }
    }

    println!("{}", dp[s.len()][2]);
    //print_dp(&dp);
}