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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))

const int mod=1000000007;
int modPow(int r,int n){
  int t=1;
  int s=r;
  while(n>0){
    if(n&0x01) t=(int64)t*s%mod;
    s=(int64)s*s%mod;
    n>>=1;
  }
  return t;
}

int *dp=NULL;
int calc(char *t,char *s,int n){
  int *dp2=dp+2*(n+1);
  int *dp1=dp+1*(n+1);
  int *dp0=dp+0*(n+1);
  int i;
  for(i=0;i<=n;i++){
    dp2[i]=dp1[i]=dp0[i]=0;
  }
  for(i=n-1;i>=0;i--){
    dp2[i]=dp2[i+1];
    if(s[i]==t[2]) dp2[i]++;
  }
  for(i=n-1;i>=0;i--){
    dp1[i]=dp1[i+1];
    if(s[i]==t[1]){
      dp1[i]+=dp2[i+1];
      dp1[i]%=mod;
    }
  }
  for(i=n-1;i>=0;i--){
    dp0[i]=dp0[i+1];
    if(s[i]==t[0]){
      dp0[i]+=dp1[i+1];
      dp0[i]%=mod;
    }      
  }
  return dp0[0];
}

void run(void){
  char *s=(char *)malloc(sizeof(char)*(100000+1));
  scanf("%s",s);
  int n=0;
  while(s[n]!='\0') n++;
  dp=(int *)malloc(sizeof(int)*3*(n+1));
  int cnt=0;
  int i;
  for(i=0;i<n;i++) if(s[i]=='?') cnt++;
  int64 ans=0;
  ans+=(int64)calc("ABC",s,n)*modPow(3,cnt);
  ans+=(int64)calc("AB?",s,n)*modPow(3,cnt-1);
  ans+=(int64)calc("A?C",s,n)*modPow(3,cnt-1);
  ans+=(int64)calc("?BC",s,n)*modPow(3,cnt-1);
  ans+=(int64)calc("A??",s,n)*modPow(3,cnt-2);
  ans+=(int64)calc("?B?",s,n)*modPow(3,cnt-2);
  ans+=(int64)calc("??C",s,n)*modPow(3,cnt-2);
  ans+=(int64)calc("???",s,n)*modPow(3,cnt-3);
  printf("%lld\n",ans%mod);
}

int main(void){
  run();
  return 0;
}