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 Python (3.4.3) (judge verdict: Accepted)

S=input()
Mod=10**9+7
dp=[1,0,0,0]

for s in S:
  if s=="A":
#dp[1]:="A"ができる個数。
    dp[1]+=dp[0]
  elif s=="B":
#dp[2]:="AB"ができる個数="B"が出てきたらこれまでの"A"個数分の"AB"ができる
    dp[2]+=dp[1]
  elif s=="C":
#dp[3]:="ABC"ができる個数="C"が出てきたらこれまでの"AB"個数分の"ABC"ができる
    dp[3]+=dp[2]
  else:
#"?"が登場した場合。"A""B""C"それぞれになるので、dp[0]~dp[3]全てが３通り（=3倍）増える。
#加えて"C"になる場合を考えると、"ABC"になる場合は"AB"の個数分増える。以下"AB""A"も同様。
    dp[3]=dp[3]*3+dp[2]
    dp[2]=dp[2]*3+dp[1]
    dp[1]=dp[1]*3+dp[0]
    dp[0]*=3
#modをとる。
  for i in range(4):
    dp[i]%=Mod

#最終的に"ABC"になる場合の数を示すdp[3]を出力。
print(dp[3])