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_macros)]
macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut iter = $ s . split_whitespace ( ) ; let mut next = || { iter . next ( ) . unwrap ( ) } ; input_inner ! { next , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { let stdin = std :: io :: stdin ( ) ; let mut bytes = std :: io :: Read :: bytes ( std :: io :: BufReader :: new ( stdin . lock ( ) ) ) ; let mut next = move || -> String { bytes . by_ref ( ) . map ( | r | r . unwrap ( ) as char ) . skip_while ( | c | c . is_whitespace ( ) ) . take_while ( | c |! c . is_whitespace ( ) ) . collect ( ) } ; input_inner ! { next , $ ( $ r ) * } } ; }
#[allow(unused_macros)]
macro_rules ! input_inner { ( $ next : expr ) => { } ; ( $ next : expr , ) => { } ; ( $ next : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ next , $ t ) ; input_inner ! { $ next $ ( $ r ) * } } ; }
#[allow(unused_macros)]
macro_rules ! read_value { ( $ next : expr , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ next , $ t ) ) ,* ) } ; ( $ next : expr , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ next , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ next : expr , chars ) => { read_value ! ( $ next , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ next : expr , usize1 ) => { read_value ! ( $ next , usize ) - 1 } ; ( $ next : expr , $ t : ty ) => { $ next ( ) . parse ::<$ t > ( ) . expect ( "Parse error" ) } ; }

#[allow(unused_macros)]
macro_rules ! vec_md { ( $ init : expr ; $ ( $ dim : expr ) ;* ) => { vec_md ! ( $ ( $ dim ) ;* => $ init ) } ; ( $ head : expr ; $ ( $ tail : expr ) ;* => $ init : expr ) => { vec ! [ vec_md ! ( $ ( $ tail ) ;* => $ init ) ; $ head ] } ; ( $ head : expr => $ init : expr ) => { vec ! [ $ init ; $ head ] } ; }

#[allow(dead_code)]
pub const M: u64 = 1_000_000_007;

fn main() {
    input! {
        s: chars
    }
    let n = s.len();
    let mut dp = vec_md![0u64;n+1;4];
    dp[0][0] = 1;
    for i in 0..n {
        let k = if s[i] == '?' {3} else {1};
        for j in 0..4 {
            dp[i+1][j] = dp[i][j] * k;
            dp[i+1][j] %= M;
        }
        match s[i] {
            '?' => {
                for j in 0..3 {
                    dp[i+1][j+1] += dp[i][j]; 
                    dp[i+1][j+1] %= M;
                }
            },
            _ => {
                let x = (s[i] as u8 - b'A') as usize;
                dp[i+1][x+1] += dp[i][x];
                dp[i+1][x+1] %= M;
            }
        }
    }
    println!("{}", dp[n][3]);
}