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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
  int H,W;
  scanf("%d%d",&H,&W);
  int a[H+1][W+1];
  int i,j;
  for(i=1;i<=H;i++){
    for(j=1;j<=W;j++){
      scanf("%d", &a[i][j]);
      a[i][j]=a[i][j]%2;
    }
  }
  int ans[H*W][4];
  int k=0;
 
  for(i=1;i<=H;i++){
    if(i%2==1){
      for(j=1;j<=W;j++){
	if(a[i][j]==1){
	  if(j==W && i!=H){
	    a[i+1][W]=(a[i+1][W]+1)%2;
	    ans[k][0]=i;ans[k][1]=W;ans[k][2]=i+1;ans[k][3]=W;
	    k=k+1;
	  }
	  if(j!=W){
	    a[i][j+1]=(a[i][j+1]+1)%2;
	    ans[k][0]=i;ans[k][1]=j;ans[k][2]=i;ans[k][3]=j+1;
	    k=k+1;
	  }
	}
      }
    }
    if(i%2==0){
      for(j=W;1<=j;j--){
        if(a[i][j]==1){
          if(j==1 && i!=H){
            a[i+1][1]=(a[i+1][1]+1)%2;
            ans[k][0]=i;ans[k][1]=1;ans[k][2]=i+1;ans[k][3]=1;
            k=k+1;
          }
	  if(j!=1){
	    a[i][j-1]=(a[i][j-1]+1)%2;
	    ans[k][0]=i;ans[k][1]=j;ans[k][2]=i;ans[k][3]=j-1;
            k=k+1;
          }
        }
      }
    }
  }

  printf("%d\n",k);
  for(i=0;i<k;i++){
    printf("%d %d %d %d\n",ans[i][0], ans[i][1],ans[i][2],ans[i][3]);
  }
  return 0;
}