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>
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rrep(i, n) for(int i = ((int)(n)-1); i >= 0; i--)
#define all(x) (x).begin(),(x).end()
 
using namespace std;
using ll = long long;
 
struct edge { int to, cost; };
 
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> P;
 
const int INF = ~(1<<31);
const int MOD = 1e9+7;
const double EPS = 1e-10;
const double PI = acos(-1.0);
 
const int mx4[] = {0,1,0,-1};
const int my4[] = {1,0,-1,0};
const int mx8[] = {0,1,1,1,0,-1,-1,-1};
const int my8[] = {1,1,0,-1,-1,-1,0,1};

ll dp[100010][4];

int main(){
	string s;
	cin >> s;

	dp[0][0] = 1;
	for(int i = 0; i < s.size(); i++){
		for(int j = 0; j < 4; j++){
			dp[i+1][j] += dp[i][j] * (s[i] == '?' ? 3 : 1);
			dp[i+1][j] %= MOD;
		}
		for(int j = 0; j < 3; j++){
			if(s[i] == '?' || s[i] == 'A'+j){
				dp[i+1][j+1] += dp[i][j];
				dp[i+1][j+1] %= MOD;
			}
		}
	}

	cout << dp[s.size()][3];

	return 0;
}