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.

Examples

Input

A??C


Output

8


Input

ABCBC


Output

3


Input

????C?????B??????A???????


Output

979596887
#include<cstring>
#include<iostream>

using namespace std;

typedef long long ll;

#define MOD 1000000007

string s;
ll dp[100001][4];

ll solve(int p, int c){
	if(dp[p][c] != -1) return dp[p][c];
	if(p == s.length() && c == 3) return dp[p][c] = 1;
	if(p == s.length() && c != 3) return dp[p][c] = 0;
	ll res = solve(p + 1, c);
	if(s[p] == '?') res *= 3;
	if(c < 3 && (s[p] == "ABC"[c] || s[p] == '?')) res += solve(p + 1, c + 1);
	return dp[p][c] = res % MOD;
}

int main(){
	cin >> s;
	memset(dp, -1, sizeof(dp));
	cout << solve(0, 0) << endl;
	return 0;
}