Problem: This is an interactive task.

We have 2N balls arranged in a row, numbered 1, 2, 3, ..., 2N from left to right, where N is an odd number. Among them, there are N red balls and N blue balls.

While blindfolded, you are challenged to guess the color of every ball correctly, by asking at most 210 questions of the following form:

* You choose any N of the 2N balls and ask whether there are more red balls than blue balls or not among those N balls.



Now, let us begin.

Constraints

* 1 \leq N \leq 99
* N is an odd number.



* * *



Example

Input




Output

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ss(x) (int) x.size()
#define cat(x) cout << #x << " = " << x << endl
 
using namespace std;

int n;

int query1(int nr, int l, int r) {
	cout << "? " << nr << " ";
	for(int i = l; i <= r; ++i) {
		cout << i << " ";
	}
	cout << endl;
	string x;
	cin >> x;
	return x[0] == 'R';
}

int query2(int nr, int l, int r) {
	cout << "? " << nr << " ";
	for(int i = 1; i <= l; ++i)
		cout << i << " ";
	for(int i = r; i <= 2 * n; ++i)
		cout << i << " ";
	cout << endl;
	string x;
	cin >> x;
	return x[0] == 'R';
}

int ans[200];
		

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	
	int beg = query1(1, 2, n);
	
	int l = 1;
	int r = n;
	
	while(l < r) {
		int m = (l + r + 1) / 2;
		if(query1(m, m + 1, m + n - 1) == beg)
			l = m;
		else
			r = m - 1;
	}
	
	for(int i = 1; i <= l; ++i) 
		ans[i] = query1(i, l + 1, l + n - 1);
	for(int i = l + 1; i < l + n; ++i)
		ans[i] = query2(i, l - 1, l + n + 1);
	for(int i = l + n; i <= 2 * n; ++i)
		ans[i] = query1(i, l + 1, l + n - 1);
	cout << "! ";
	for(int i = 1; i <= 2 * n; ++i) 
		cout << (ans[i] == 1 ? "R" : "B");
	
		
	
	return 0;
}