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)

#![allow(unused_imports)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

pub trait InputValue {
    fn parse(s: &str) -> Self;
}

pub fn read<T: InputValue>() -> T {
    let mut buf = String::new();
    let _ = stdin().read_line(&mut buf);
    T::parse(&buf.trim())
}

pub fn readn<T: InputValue>(n: usize) -> Vec<T> {
    let mut vec = vec![];
    for _ in 0..n {
        vec.push(read());
    }
    vec
}

pub fn readnc<T: InputValue>() -> Vec<T> {
    let mut vec = vec![];
    let line: String = read();
    for token in line.split_whitespace() {
        vec.push(T::parse(token));
    }
    vec
}

macro_rules! parse_single_value {
    ($($t:ty),*) => {
        $(
            impl InputValue for $t {
                fn parse(s: &str) -> $t { s.parse().unwrap() }
            }
        )*
	}
}
parse_single_value!(i32, i64, f32, f64, usize, String);

macro_rules! parse_tuple {
	($($t:ident),*) => {
		impl<$($t),*> InputValue for ($($t),*) where $($t: InputValue),* {
			fn parse(s: &str) -> ($($t),*) {
				let mut tokens = s.split_whitespace();
				let t = ($($t::parse(tokens.next().unwrap())),*);
				t
			}
		}
	}
}
parse_tuple!(A, B);
parse_tuple!(A, B, C);


// ===

const MOD: i64 = 1000000007;

fn main() {
    let s: String = read();
    let s: Vec<u8> = s.into_bytes();
    let n = s.len();

    let mut pow3 = vec![1; n+10];
    for i in 1..pow3.len() {
        pow3[i] = pow3[i-1] * 3 % MOD;
    }

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

    let mut prev = vec![vec![0; 4]; n+1];
    let mut next = vec![vec![0; 4]; n+1];
    for i in 0..n {
        for j in 0..4 {
            let pa = if s[i] == abc[j] { 1 } else { 0 };
            let pb = if s[n-1-i] == abc[j] { 1 } else { 0 };
            prev[i+1][j] += prev[i][j] + pa;
            next[i+1][j] += next[i][j] + pb;
        }
    }

    let mut sum = 0;
    for i in 1..n-1 {
        let mut sharp = prev[n][3] as usize;
        if s[i] == b'B' {
        } else if s[i] == b'?' {
            sharp -= 1;
        } else {
            continue;
        }
        let aa = prev[i][0];
        let ax = prev[i][3];
        let cc = next[n-1-i][2];
        let cx = next[n-1-i][3];
        let mut add = 0;

        // *_*
        if ax >= 1 && cx >= 1 {
            add += ax * cx % MOD * pow3[sharp-2] % MOD;
        }

        // a_* or *_c
        if ax >= 1 || cx >= 1 {
            add += ((aa * cx) + (ax * cc)) % MOD * pow3[sharp-1] % MOD;
        }

        // a_c
        add += aa * cc % MOD * pow3[sharp] % MOD;

        sum += add;
        sum %= MOD;
    }

    println!("{}", sum);
}