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(){
  long long i,c=1,M=1e18+7,d[100010][3]={};
  char s[100010];
  scanf("%s",s);
  for(i=0;s[i];i++){
    d[i+1][0]=d[i][0];
    d[i+1][1]=d[i][1];
    d[i+1][2]=d[i][2];
    if(s[i]=='A')d[i+1][0]=(d[i+1][0]+c      )%M;
    if(s[i]=='B')d[i+1][1]=(d[i+1][1]+d[i][0])%M;
    if(s[i]=='C')d[i+1][2]=(d[i+1][2]+d[i][1])%M;
    if(s[i]=='?'){
      d[i+1][0]=(d[i+1][0]*3+c      )%M;
      d[i+1][1]=(d[i+1][1]*3+d[i][0])%M;
      d[i+1][2]=(d[i+1][2]*3+d[i][1])%M;
      c=(c*3)%M;
    }

      
    //if(s[i]=='A'||s[i]=='?')d[i+1][0]+=(+1      )%M;
    //if(s[i]=='B'||s[i]=='?')d[i+1][1]+=(d[i][0]+d[i][1]*3)%M;
    //if(s[i]=='C'||s[i]=='?')d[i+1][2]+=(d[i][1]+d[i][2]*3)%M;
  }
  printf("%lld\n",d[i][2]);
  for(i=1;s[i-1];i++)printf("%lld %lld %lld\n",d[i][0],d[i][1],d[i][2]);
  return 0;
}