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 <bits/stdc++.h>
using namespace std;
#define REP(i,N) for(i=0;i<N;i++)
#define INF 1e9
#define MOD 1000000007 
typedef long long ll;

int main(void){
  string S;cin >> S;
  int N=S.length();
  map<char,int> m = {
    {'A',0},
    {'B',1},
    {'C',2},
  };
  //
  int i,j;
  ll dp[N+1][4];
  //初期化
  REP(j,4)dp[N][j] = 0;
  dp[N][3] = 1;
  for(i=N;i>=0;i--){
    for(j=3; j>=0; j--){
      if(i==N) dp[i][j] = j==3 ? 1:0;
      else{
        dp[i][j] = dp[i+1][j] * (S[i] == '?' ? 3 : 1 );
        if(j<3 && ( S[i] == '?' || j == m[S[i]] ) ) dp[i][j] += dp[i+1][j+1];
        dp[i][j] %= MOD;
      }
    }
  }
  cout << dp[0][0] << endl;;
  return 0;
}
