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::str::FromStr;

fn mod_pow(x:i64, n: i64, m:i64) -> i64 {
    if n <= 0 {
        return 1_i64
    }
    if x <= 0 {
        return 1_i64
    }
    if n % 2 == 1 {
        return (x * mod_pow(x, n-1, m)) % m
    }
    return mod_pow((x*x) % m, n/2, m)
}

fn main(){
    let po = 10_i64.pow(9) + 7;

    let s: Vec<char> = read::<String>().chars().collect();
    let len = s.len();
    
    let mut ta: Vec<i64> = vec![0;len+1];
    let mut tc: Vec<i64> = vec![0;len+1];
    let mut tq1: Vec<i64> = vec![0;len+1];
    let mut tq2: Vec<i64> = vec![0;len+1];
    for i in 1..(len+1) {
        match s[i-1] {
            'A' => ta[i] += 1,
            'C' => tc[i-1] += 1,
            '?' => {
                tq1[i] += 1;
                tq2[i-1] += 1;
            },
            _ => {},
        }
    }
    for i in 1..(len+1) {
        ta[i] += ta[i-1];
        tc[len-i] += tc[len-i+1];
        tq1[i] += tq1[i-1];
        tq2[len-i] += tq2[len-i+1];
    }

    let mut ans: i64 = 0;
    for i in 1..len {
        if s[i] == 'B' || s[i] == '?' {
            let a1 = (ta[i] * mod_pow(3, tq1[i], po)) % po;
            let a2 = 
                if tq1[i] > 0 {
                    (tq1[i] * mod_pow(3, tq1[i]-1, po)) % po
                } else {0};
            let b1 = (tc[i+1] * mod_pow(3, tq2[i+1], po)) % po;
            let b2 = 
                if tq2[i+1] > 0 {
                    (tq2[i+1] * mod_pow(3, tq2[i+1]-1, po)) % po
                } else {0};
            ans += (a1 + a2) * (b1 + b2);
            ans %= po;
        }
    }

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

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char) 
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}