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)

const MOD: u64 = 1000000007;

fn main() {
	let mut s = String::new();
	std::io::stdin().read_line(&mut s).unwrap();
	let s: Vec<_> = s.trim().bytes().collect();
	let mut cum = vec![(0u32, 0u32, 0u32, 0u32); s.len() + 1];
	for (i, &b) in s.iter().enumerate() {
		match b {
			b'A' => cum[i + 1] = (cum[i].0 + 1, cum[i].1, cum[i].2, cum[i].3),
			b'B' => cum[i + 1] = (cum[i].0, cum[i].1 + 1, cum[i].2, cum[i].3),
			b'C' => cum[i + 1] = (cum[i].0, cum[i].1, cum[i].2 + 1, cum[i].3),
			_    => cum[i + 1] = (cum[i].0, cum[i].1, cum[i].2, cum[i].3 + 1),

		}
	}
	let mut ans = 0;
	for (i, &b) in s[1..s.len() - 1].iter().enumerate() {
		if b == b'A' || b == b'C' { continue; }
		let al = cum[i + 1].0;
		let ql = cum[i + 1].3;
		let cr = cum[s.len()].2 - cum[i + 2].2;
		let qr = cum[s.len()].3 - cum[i + 2].3;
		ans = (ans +
			((al as u64) * mod_pow(3, ql) + if ql > 0 { (ql as u64) * mod_pow(3, ql - 1) } else { 0 }) % MOD *
			(((cr as u64) * mod_pow(3, qr) + if qr > 0 { (qr as u64) * mod_pow(3, qr - 1) } else { 0 }) % MOD)) % MOD;
	}
	println!("{}", ans);
}

fn mod_pow(a: u64, n: u32) -> u64 {
	let mut r = 1;
	let mut a = a;
	let mut n = n;
	while n > 0 {
		if n & 1 != 0 {
			r = r * a % MOD;
		}
		a = a * a % MOD;
		n >>= 1;
	}
	r
}