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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
int main(){
  string s;
  cin >> s;
  int n=s.size();
  vector<long long> A(n+1,0),B(n+1,0),C(n+1,0),D(n+1,0);
  D[0]=1;
  for(int i=0;i<n;i++){
    A[i+1]=A[i];
    C[i+1]=C[i];
    B[i+1]=B[i];
    D[i+1]=D[i];
    if(s[i]=='A')
      A[i+1]=(A[i+1]+D[i])%(1000000007);
    else if(s[i]=='B')
      B[i+1]=(B[i+1]+A[i])%(1000000007);
    else if(s[i]=='C')
      C[i+1]=(C[i+1]+B[i])%(1000000007);
    else{
      A[i+1]=(long long)(A[i+1]*3+D[i])%(1000000007);
      B[i+1]=(long long)(B[i+1]*3+A[i])%(1000000007);
      C[i+1]=(long long)(C[i+1]*3+B[i])%(1000000007);
      D[i+1]=(long long)(D[i+1]*3)%(1000000007);
    }
  }
  cout << C[n] << endl;
 // for(int i=0;i<n+1;i++)
   // cout << D[i] <<' ' << A[i] << ' ' << B[i] << ' ' << C[i] << endl;
}