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::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

#[allow(dead_code)]
fn with_stdout_lock<F: FnOnce(std::io::BufWriter<std::io::StdoutLock>)>(f: F) {
    let out = std::io::stdout();
    let w = std::io::BufWriter::new(out.lock());
    f(w);
}

#[allow(unused_macros)]
macro_rules! debug {
    ($var:expr) => {
        match &$var { v => eprintln!(concat!(stringify!($var), "={:?}"), v) }
    };
    ($var:expr, $($vars:expr),*) => {
        match &$var { v => eprint!(concat!(stringify!($var), "={:?} "), v) }
        debug!($($vars),*);
    };
}

fn main() {
    let s: Vec<char> = input::<String>().chars().collect();
    const MOD: i64 = 1000000007;
    const Z: usize = 0;
    const A: usize = 1;
    const AB: usize = 2;
    const ABC: usize = 3;

    let mut dp = vec![vec![0; 4]; s.len() + 1];
    dp[0][0] = 1;
    for i in 0..s.len() {
        if s[i] == 'A' {
            dp[i + 1][Z] = dp[i][Z];
            dp[i + 1][A] = (dp[i][A] + dp[i][Z]) % MOD;
            dp[i + 1][AB] = dp[i][AB];
            dp[i + 1][ABC] = dp[i][ABC];
        } else if s[i] == 'B' {
            dp[i + 1][Z] = dp[i][Z];
            dp[i + 1][A] = dp[i][A];
            dp[i + 1][AB] = (dp[i][AB] + dp[i][A]) % MOD;
            dp[i + 1][ABC] = dp[i][ABC];
        } else if s[i] == 'C' {
            dp[i + 1][Z] = dp[i][Z];
            dp[i + 1][A] = dp[i][A];
            dp[i + 1][AB] = dp[i][AB];
            dp[i + 1][ABC] = (dp[i][ABC] + dp[i][AB]) % MOD;
        } else {
            dp[i + 1][Z] = (3 * dp[i][Z]) % MOD;
            dp[i + 1][A] = (3 * dp[i][A] + dp[i][Z]) % MOD;
            dp[i + 1][AB] = (3 * dp[i][AB] + dp[i][A]) % MOD;
            dp[i + 1][ABC] = (3 * dp[i][ABC] + dp[i][AB]) % MOD;
        }
    }

    println!("{}", dp[s.len()][ABC]);
}