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 Scala (2.11.7) (judge verdict: Accepted)

import java.io._
import java.util.StringTokenizer

import scala.collection.mutable
import scala.util.Sorting
import math.{abs, max, min}
import scala.collection.mutable.{ArrayBuffer, ListBuffer}
import scala.reflect.ClassTag

object Main {
  val MOD = 1000000007
  val out = new PrintWriter(System.out)

  def solve(): Unit = {
    val sc = new InputReader(System.in)
    val S = sc.next()

    val dp = Array.ofDim[Long](S.length + 1, 4)
    dp(0)(0) = 1
    1 to S.length foreach { i =>
      rep(4) { j =>
        if (S(i - 1) == '?' && j == 0) {
          dp(i)(j) = dp(i - 1)(j) * 3 % MOD
        } else if (S(i - 1) == '?' && j > 0) {
          dp(i)(j) = (dp(i - 1)(j) * 3 + dp(i - 1)(j - 1)) % MOD
        } else if (j == (S(i - 1) - 'A' + 1)) {
          dp(i)(j) = (dp(i - 1)(j) + dp(i - 1)(j - 1)) % MOD
        } else {
          dp(i)(j) = dp(i - 1)(j)
        }
      }
    }
    val ans = dp(S.length)(3)

    out.println(ans)
  }

  def main(args: Array[String]): Unit = {
    solve()
    out.flush()
  }


  class InputReader(val stream: InputStream) {
    private val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private var tokenizer: StringTokenizer = _

    def next(): String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  def rep(n: Int)(f: Int => Unit): Unit = {
    var i = 0
    while(i < n) { f(i); i += 1 }
  }
  def rep_r(n: Int)(f: Int => Unit): Unit = {
    var i = n - 1
    while(i >= 0) { f(i); i -= 1 }
  }

  def map[@specialized A: ClassTag](n: Int)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    rep(n)(i => res(i) = f(i))
    res
  }

  implicit class ArrayOpts[A](val as: Array[A]) extends AnyVal {
    // todo Orderingだとboxing発生するので自作Orderを用意したい
    def maxByOpt[B: Ordering](f: A => B): Option[A] = {
      if (as.nonEmpty) Some(as.maxBy(f)) else None
    }

    def grpBy[K](f: A => K): mutable.Map[K, ArrayBuffer[A]] = {
      val map = mutable.Map.empty[K, ArrayBuffer[A]]
      rep(as.length)(i => map.getOrElseUpdate(f(as(i)), ArrayBuffer()) += as(i))
      map
    }

    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      var sum = num.zero
      rep(as.length)(i => sum = num.plus(sum, f(as(i))))
      sum
    }

    def minByEx[B](f: A => B, ixRange: Range = as.indices)(implicit cmp: Ordering[B]): (A, B) = {
      limit(f, ixRange)(cmp.lt)
    }

    def maxByEx[B](f: A => B, ixRange: Range = as.indices)(implicit cmp: Ordering[B]): (A, B) = {
      limit(f, ixRange)(cmp.gt)
    }

    private def limit[B](f: A => B, ixRange: Range)(cmp: (B, B) => Boolean): (A, B) = {
      var limA = as(ixRange.head)
      var limB = f(limA)

      for (i <- ixRange.tail) {
        val a = as(i)
        val b = f(a)
        if (cmp(b, limB)) {
          limA = a
          limB = b
        }
      }
      (limA, limB)
    }
  }

  implicit class IterableOpts[A](val as: Iterable[A]) extends AnyVal {
    def sumBy[B](f: A => B)(implicit num: Numeric[B]): B = {
      as.foldLeft(num.zero)((acc, a) => num.plus(acc, f(a)))
    }
  }
}