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;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const LL mod = 1e9 + 7;
const int MC = 1e5 + 5;

LL a[MC],c[MC],qa[MC],qc[MC];

int main(){
	string S; cin >> S;
	int n = S.size();
	a[0] = 0; qa[0] = 1;
	repp(i,0,n){
		if(S[i] == 'A'){
			a[i+1] = (a[i] + qa[i]) % mod;
			qa[i+1] = qa[i];
		} else if(S[i] == '?'){
			a[i+1] = (a[i] * 3 + qa[i]) % mod;
			qa[i+1] = qa[i] * 3 % mod;
		} else {
			a[i+1] = a[i];
			qa[i+1] = qa[i];
		}
	}
	c[n] = 0; qc[n] = 1;
	repm(i,n,0){
		if(S[i-1] == 'C'){
			c[i-1] = (c[i] + qc[i]) % mod;
			qc[i-1] = qc[i];
		} else if(S[i-1] == '?'){
			c[i-1] = (c[i] * 3 + qc[i]) % mod;
			qc[i-1] = qc[i] * 3 % mod;
		} else {
			c[i-1] = c[i];
			qc[i-1] = qc[i];
		}
	}
	LL ans = 0;
	repp(i,1,n-1){
		if(S[i] == 'A' || S[i] == 'C') continue;
		ans = (ans + a[i] * c[i+1]) % mod;
	}
	cout << ans << endl;
	return 0;
}