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: Wrong Answer)

#include<iostream>
#include<algorithm>

using namespace std;

#define INF (1<<30)
#define MOD 1000000009
#define ll long long

ll memo[100001][10][10];

string s;
int n;

ll dfs(int depth, int phase, int now){
	// cout << depth << " " << phase << " " << now << endl;
	if(memo[depth][phase][now]!=-1)return memo[depth][phase][now];
	if(depth == n && phase == 3)return 1;
	else if(depth == n) return 0;
	ll ans = 0;
	if(s[depth] == '?' && now == 0){
		ans+=dfs(depth, phase, 1);
		ans+=dfs(depth, phase, 2);
		ans+=dfs(depth, phase, 3);
		return memo[depth][phase][now] = ans%MOD;
	}
	if(now == 0)now = (int)s[depth];
	if(phase+1 == now){
		ans += dfs(depth+1, phase+1, 0);
	}
	ans += dfs(depth+1, phase, 0);
	return memo[depth][phase][now] = ans%MOD;
}

int main(){

	cin >> s;
	n = s.length();
	for(int i = 0;i < n;i++){
		char c = s[i];
		if(c != '?'){
			s[i] = c - 'A' + 1;
		}
	}
	fill_n(**memo,100001*10*10,-1);
	cout << dfs(0,0,0) << endl;

	return 0;
}