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)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Formatter};
use std::io::*;
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

// -----------------------------------------------
// Framework
// -----------------------------------------------

#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($t:ty) =>
        (rl().parse::<$t>().unwrap());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($arg:expr),*) => {
        #[cfg(debug_assertions)]
        {
            let entries = [$(&stringify!([$arg]:), &$arg as &Debug),*];
            stderr().write_fmt(format_args!("{:?}\n", entries)).unwrap();
        }
    };
}

#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim_right().to_owned()
}

trait IteratorExt: Iterator + Sized {
    fn vec(self) -> Vec<Self::Item> {
        self.collect()
    }
}

impl<T: Iterator> IteratorExt for T {}

// -----------------------------------------------
// Solution
// -----------------------------------------------

pub fn main() {
    let S = rl().chars().vec();
    let N = S.len();

    let P = 1_000_000_007_i64;
    let ABC = "ABC".chars().vec();

    // automaton DP?
    // dp[r][i] = t mod P
    // i : S の範囲 S[..i]
    // r : 固定した位置の個数
    // t : 範囲 S[..i] において決められた r 個の位置の文字がそれぞれ A, B, C であるような文字列の個数 (mod P)
    let mut dp = vec![vec![0; N + 1]; 3 + 1];
    dp[0][0] = 1_i64;

    for i in 0..N {
        // 位置 i をABC数に含めないケースの状態遷移
        let f = if S[i] == '?' { 3 } else { 1 };
        for r in 0..4 {
            dp[r][i + 1] = (dp[r][i + 1] + f * dp[r][i] % P) % P;
        }

        // 位置 i をABC数に含めるケースの状態遷移
        for r in 0..3 {
            if S[i] == ABC[r] || S[i] == '?' {
                dp[r + 1][i + 1] = (dp[r + 1][i + 1] + dp[r][i]) % P;
            }
        }
    }

    println!("{}", dp[3][N]);
}