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

#define MAX_LEN (100000)
#define NUM_MOD (1000000007L)

int main() {
  // read inputs
  int N;
  char cs[MAX_LEN + 1];
  scanf("%s", cs);
  N = strlen(cs);

  // calculate powers of 3
  long pow3s[MAX_LEN + 1] = {};
  pow3s[0] = 1;
  for (int i = 0; i < N; i++) {
    pow3s[i + 1] = 3 * pow3s[i] % NUM_MOD;
  }

  // solve
  long cnts[3][4] = {}, cntQ = 0;
  for (int i = 0; i < N; i++) {
    switch (cs[i]) {
      case 'A':
        cnts[0][0] = (cnts[0][0] + 1) % NUM_MOD;
        break;
      case 'B':
        for (int j = 0; j < 4; j++) {
          cnts[1][j] = (cnts[1][j] + cnts[0][j]) % NUM_MOD;
        }
        break;
      case 'C':
        for (int j = 0; j < 4; j++) {
          cnts[2][j] = (cnts[2][j] + cnts[1][j]) % NUM_MOD;
        }
        break;
      case '?':
        cntQ++;
        for (int j = 1; j < 4; j++) {
          cnts[2][j] = (cnts[2][j] + cnts[1][j - 1]) % NUM_MOD;
        }
        for (int j = 1; j < 4; j++) {
          cnts[1][j] = (cnts[1][j] + cnts[0][j - 1]) % NUM_MOD;
        }
        cnts[0][1] = (cnts[0][1] + 1) % NUM_MOD;
        break;
    }
  }

  long ans = 0;
  for (int j = 0; j < 4; j++) {
    ans = (ans + pow3s[cntQ - j] * cnts[2][j]) % NUM_MOD;
  }
  printf("%ld\n", ans);
}