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;
using ll = long long;
#define rep(i,n) for(int i=0;i<n;++i)
#define rep2(i,i0,n) for(int i=i0;i<n;++i)

int main()
{
    const ll m = 1e9+7;
    string s;
    cin >> s;
    int n = s.size();

    vector<vector<ll>> dp(n+1, vector<ll>(4, 0));
    dp[0][0] = 1;

    rep(i,n) {
        rep(j,4) dp[i+1][j] = s[i] == '?' ? (3*dp[i][j]) % m : dp[i][j];
        if(s[i] == 'A' || s[i] == '?') dp[i+1][1] = (dp[i+1][1] + dp[i][0]) % m;
        if(s[i] == 'B' || s[i] == '?') dp[i+1][2] = (dp[i+1][2] + dp[i][1]) % m;
        if(s[i] == 'C' || s[i] == '?') dp[i+1][3] = (dp[i+1][3] + dp[i][2]) % m;
    }

    cout << dp[n][3] << endl;
    return 0;
}