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>
using namespace std;
#define int long long
#define f first
#define s second
typedef pair<int,int> pi;
#define FAST ios_base::sync_with_stdio(0); cin.tie(0);
#define all(x) x.begin(),x.end()
const int maxn = 210;

int n;
int ans[maxn];

int query(int x) {
	cout << "?";
	for (int i =x;i<x+n;i++) {
		cout << " " << i;
	}
	cout << "\n"; cout.flush();
	string res; cin >> res;
	if (res == "Red") return 1;
	else return 0;
}

int test(vector <int> v, int x) {
	cout << "?";
	for (auto i: v) {
		cout << " " << i;
	}
	cout << " " << x << "\n"; cout.flush();
	string res; cin >> res;
	if (res == "Red") return 1;
	else return 0;
}

int32_t main() {
	cin >> n;
	
	int prev = query(1);
	int high = 1 + n;
	int low = 1;
	
	while (high - low > 1) {
		int mid = (high + low) / 2;
		int res = query(mid); 
		if (res == prev) low = mid;
		else high = mid;
	}
	
	ans[low] = prev;
	ans[low + n] = !prev;
	
	vector <int> v1,v2;
	for (int i = low+1; i < low + n; i++) {
		v1.push_back(i);
	}
	
	for (int i = 1; i <= low-1;i++) {
		ans[i] = test(v1,i);
		v2.push_back(i);
	}
	
	for (int i = low + n + 1; i <= n*2; i++) {
		ans[i] = test(v1, i);
		v2.push_back(i);
	}
	
	for (int i = low+1; i < low + n;i++) {
		ans[i] = test(v2,i);
	}
	cout << "! ";
	for (int i =1;i<=2*n;i++) {
		if (ans[i]) cout << "R";
		else cout << "B";
	}
  cout.flush();
}