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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program test
implicit none

character(len=100000) :: s
integer(8) :: l,temp,ans=0,big=1000000007,ruijyo(0:100000)
integer(8) :: i,j,k,a(100000)=0,b(100000)=0,c(100000)=0,q(100000)=0
integer(8) :: ax(0:100001)=0, qx(0:100001)=0
integer(8) :: cy(0:100001)=0, qy(0:100001)=0

read(*,*) s

ruijyo(0)=1
do i=1,100000
	ruijyo(i) = mod(ruijyo(i-1)*3,big)
enddo


l = len_trim(s)

do i=1,l
	if(s(i:i)=="A") then
		a(i) = 1
	else if(s(i:i)=="B") then
		b(i) = 1
	else if(s(i:i)=="C") then
		c(i) = 1
	else
		q(i) = 1
	endif
enddo

!累積
do i=1,l
	ax(i) = ax(i-1) + a(i)
	qx(i) = qx(i-1) + q(i)
enddo

do i=l,1,-1
	cy(i) = cy(i+1) + c(i)
	qy(i) = qy(i+1) + q(i)
enddo

do i=2,l-1
	if(b(i)==1 .or. q(i)==1) then
		temp = 0
		temp = temp + (ax(i-1))*(cy(i+1))*ruijyo(qx(i-1)+qy(i+1))
		temp = temp + (qx(i-1))*(cy(i+1))*ruijyo(max(0,qx(i-1)+qy(i+1)-1))
		temp = temp + (ax(i-1))*(qy(i+1))*ruijyo(max(0,qx(i-1)+qy(i+1)-1))
		temp = temp + (qx(i-1))*(qy(i+1))*ruijyo(max(0,qx(i-1)+qy(i+1)-2))
		ans = ans + mod(temp,big)
	endif
enddo

write(*,*) mod(ans,big)


end program