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)

extern crate core;

use std::io::{self, Read};

#[macro_use]
mod parser {
    macro_rules! input {
    ($s:expr=>$($t:tt)*) => {
        let mut lines=$s.split("\n");
        $(
            line_parse!(lines,$t);
        )*
    };
    }

    macro_rules! line_parse {
    ($lines:expr,($($name:ident:$t:tt)*)) => {
        let mut line=$lines.next().unwrap().split_whitespace();
        $(value_def!(line,$name,$t);)*
    };

    //複数行
    ($lines:expr,{$n:expr;$name:ident:$t:tt}) => {
        values_def!($lines,$n,$name,$t);
    };
    }

    macro_rules! value_def {
        ($line:expr, $name:ident, $t:tt) => {
            let $name = value!($line, $t);
        };
    }

    macro_rules! values_def {
        ($lines:expr, $n:expr, $name:ident, $t:tt) => {
            let $name = {
                let mut vec = Vec::new();
                for i in 0..$n {
                    let mut next = $lines.next().unwrap().split_whitespace();
                    vec.push(value!(next, $t));
                }
                vec
            };
        };
    }

    macro_rules! value {
    //配列
    ($line:expr,[$t:tt]) => {
        $line.map(|x|{
        let mut iter=::std::iter::once(x);
        value!(iter,$t)
        }).collect::<Vec<_>>()
    };
    //タプル
    ($line:expr,($($t:tt),*)) => {
        ($(value!($line,$t),)*)
    };
    //文字列
    ($line:expr,#) => {
        $line.next().unwrap()
    };
    //単一値
    ($line:expr,$t:ty) => {
        $line.next().unwrap().parse::<$t>().unwrap()
    };
    }
}

fn main() {
    let mut input = String::new();
    io::stdin().read_to_string(&mut input).unwrap();
    let output = solve(input.trim().to_string());
    println!("{}", output);
}

#[derive(Debug, Clone, Copy, Eq, PartialEq)]
enum Type {
    A,
    B,
    C,
    Q,
}

fn solve(input: String) -> String {
    input!(input=>(s:#));
    let mut list = s
        .chars()
        .map(|x| match x {
            'A' => Type::A,
            'B' => Type::B,
            'C' => Type::C,
            _ => Type::Q,
        })
        .collect::<Vec<_>>();

    list.reverse();

    let q_count = list.clone().into_iter().filter(|&x| x == Type::Q).count();
    let mut q_count_pow_mod = 1;
    for _ in 0..q_count {
        q_count_pow_mod = mul(q_count_pow_mod, 3);
    }

    let mut a = 0i64;
    let mut b = 0i64;
    let mut c = 0i64;
    for x in list {
        match x {
            Type::A => a = add(a, b),
            Type::B => b = add(b, c),
            Type::C => c = add(c, q_count_pow_mod),
            Type::Q => {
                a += div(b, 3);
                b += div(c, 3);
                c += div(q_count_pow_mod, 3);
            }
        }
    }
    (a % 1000000007).to_string()
}

fn power(x: i64, y: i64) -> i64 {
    if y == 0 {
        1
    } else if y == 1 {
        x % 1000000007
    } else if y % 2 == 0 {
        power(x, y / 2).pow(2) % 1000000007
    } else {
        power(x, y / 2).pow(2) * x % 1000000007
    }
}

fn div(a: i64, b: i64) -> i64 {
    mul(a, power(b, 1000000007 - 2))
}

fn add(a: i64, b: i64) -> i64 {
    (a + b) % 1000000007
}

fn mul(a: i64, b: i64) -> i64 {
    ((a % 1000000007) * (b % 1000000007)) % 1000000007
}

macro_rules! tests {
    ($($name:ident: $input:expr=>$output:expr,)*) => {
        mod tests {
            $(
                #[test]
                fn $name() {
                    assert_eq!($output.trim().to_string(),super::solve($input.trim().to_string()));
                }
            )*
        }
    }
}

tests! {
    test1: "A??C" => "8",
    test2: "ABCBC" => "3",
    test3: "????C?????B??????A???????" => "979596887",
    test_add1:"A?C"=>"1",
}