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::*;
use std::ops::*;
const MOD: u64 = 1_000_000_007;

#[derive(Debug, Copy, Clone)]
struct Mint {
    value: u64,
}

impl Add for Mint {
    type Output = Mint;
    fn add(self, other: Mint) -> Mint {
        let mut res = self.value + other.value;
        if res >= MOD {
            res -= MOD;
        }
        Mint { value: res }
    }
}
impl AddAssign for Mint {
    fn add_assign(&mut self, other: Mint) {
        let mut res = self.value + other.value;
        if res >= MOD {
            res -= MOD;
        }
        *self = Mint { value: res };
    }
}

impl std::fmt::Display for Mint {
    fn fmt(&self, f: &mut std::fmt::Formatter) -> std::fmt::Result {
        write!(f, "{}", self.value)
    }
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let s: Vec<char> = itr.next().unwrap().chars().collect();

    let n = s.len();
    let mut dp: Vec<Vec<Mint>> = vec![vec![Mint { value: 0 }; 4]; n + 1];

    dp[0][0] = Mint { value: 1 };
    for i in 0..n {
        if s[i] == 'A' || s[i] == '?' {
            // use A
            dp[i + 1][1] += dp[i][0];
            // not use
            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];
        }
        if s[i] == 'B' || s[i] == '?' {
            // use B
            dp[i + 1][2] += dp[i][1];
            // not use
            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];
        }
        if s[i] == 'C' || s[i] == '?' {
            // use C
            dp[i + 1][3] += dp[i][2];
            // not use
            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];
        }
    }
    println!("{}", dp[n][3]);
}