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 (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
use proconio::{
   fastout, input,
   marker::{Bytes, Isize1, Usize1},
};

#[fastout]
fn main() {
   input! {
      s: Bytes,
   }
   let n = s.len();
   let mut dp = vec![vec![Mint::new(0); 4]; n + 1];
   dp[n][3] += 1;
   for i in (0..n).rev() {
      if s[i] == b'?' {
         dp[i][3] = dp[i + 1][3] * 3;
         for j in 0..3 {
            dp[i][j] = dp[i + 1][j] * 3 + dp[i + 1][j + 1];
         }
      } else {
         dp[i][3] = dp[i + 1][3];
         for (j, c) in (b'A'..=b'C').enumerate() {
            dp[i][j] = dp[i + 1][j] + dp[i + 1][j + 1] * if s[i] == c { 1 } else { 0 };
         }
      }
   }
   println!("{}", dp[0][0]);
}

const MOD: i64 = 1000000007;
#[derive(Copy, Clone)]
pub struct Mint {
   x: i64,
}
impl Mint {
   fn new(x: i64) -> Self {
      Mint {
         x: (x % MOD + MOD) % MOD,
      }
   }
   fn pow(self, n: usize) -> Self {
      match n {
         0 => Mint::new(1),
         _ => {
            let mut a = self.pow(n >> 1);
            a *= a;
            if n & 1 == 1 {
               a *= self;
            }
            a
         }
      }
   }
   fn inv(self) -> Self { self.pow((MOD - 2) as usize) }
}
impl std::ops::Neg for Mint {
   type Output = Mint;
   fn neg(self) -> Self::Output { Self::new(-self.x) }
}
impl std::ops::AddAssign<Mint> for Mint {
   fn add_assign(&mut self, rhs: Self) {
      self.x += rhs.x;
      if self.x >= MOD {
         self.x -= MOD;
      }
   }
}
impl std::ops::AddAssign<i64> for Mint {
   fn add_assign(&mut self, rhs: i64) { *self += Mint::new(rhs); }
}
impl<T> std::ops::Add<T> for Mint
where
   Mint: std::ops::AddAssign<T>,
{
   type Output = Self;
   fn add(self, other: T) -> Self {
      let mut res = self;
      res += other;
      res
   }
}
impl std::ops::SubAssign<Mint> for Mint {
   fn sub_assign(&mut self, rhs: Self) {
      self.x += MOD - rhs.x;
      if self.x >= MOD {
         self.x -= MOD;
      }
   }
}
impl std::ops::SubAssign<i64> for Mint {
   fn sub_assign(&mut self, rhs: i64) { *self -= Mint::new(rhs); }
}
impl<T> std::ops::Sub<T> for Mint
where
   Mint: std::ops::SubAssign<T>,
{
   type Output = Self;
   fn sub(self, other: T) -> Self {
      let mut res = self;
      res -= other;
      res
   }
}
impl std::ops::MulAssign<Mint> for Mint {
   fn mul_assign(&mut self, rhs: Self) {
      self.x *= rhs.x;
      self.x %= MOD;
   }
}
impl std::ops::MulAssign<i64> for Mint {
   fn mul_assign(&mut self, rhs: i64) { *self *= Mint::new(rhs); }
}
impl<T> std::ops::Mul<T> for Mint
where
   Mint: std::ops::MulAssign<T>,
{
   type Output = Self;
   fn mul(self, other: T) -> Self {
      let mut res = self;
      res *= other;
      res
   }
}
impl std::ops::DivAssign<Mint> for Mint {
   fn div_assign(&mut self, rhs: Self) { *self *= rhs.inv(); }
}
impl std::ops::DivAssign<i64> for Mint {
   fn div_assign(&mut self, rhs: i64) { *self *= Mint::new(rhs).inv(); }
}
impl<T> std::ops::Div<T> for Mint
where
   Mint: std::ops::DivAssign<T>,
{
   type Output = Self;
   fn div(self, other: T) -> Self {
      let mut res = self;
      res /= other;
      res
   }
}
impl<T> From<T> for Mint
where
   i64: From<T>,
{
   fn from(x: T) -> Mint { Mint::new(i64::from(x)) }
}
impl std::fmt::Display for Mint {
   fn fmt(&self, f: &mut std::fmt::Formatter<'_>) -> std::fmt::Result { write!(f, "{}", self.x) }
}