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 fi first
#define se second
#define long long long
typedef pair<int,int> ii;

int n;
bool ans[203];
bool ins[203];

void print(vector<int> vec)
{
	for(int i = 0; i < n; i++)
		cout << vec[i] << ((i == n-1)? '\n' : ' ');
	fflush(stdout);
}


int ask(vector<int> vec)
{
	cout << "? ";
	print(vec);
	string pet;
	cin >> pet;
	return (pet == "Red")? 0 : 1;
}

vector<int> cons(int x)
{
	vector<int> vec;
	for(int i = 1; i <= n-x; i++) vec.push_back(i);
	for(int i = n+1; i <= n+x; i++) vec.push_back(i);
	return vec;
}

vector<int> binser(int bef, int &piv1, int &piv2, int &idx)
{
	int kir = 0, kan = n;
	while(kir < kan)
	{
		int mid = (1+kir+kan)/2;
		if(ask(cons(mid)) == bef) kir = mid;
		else kan = mid-1;
	}
	idx = n-kir-1;
	piv1 = n-kir;
	piv2 = n+kir+1;
	return cons(kir);

}

int main()
{
	scanf("%d", &n);
	vector<int> vec, nvec;
	for(int i = 1; i <= n; i++)
		vec.push_back(i);

	int bef = ask(vec), piv1, piv2, idx;
	vec = binser(bef, piv1, piv2, idx);

	ans[piv1] = bef;
	ans[piv2] = bef^1;
	
	for(int x : vec) ins[x] = 1;

	nvec = {piv2};
	for(int i = 1; i <= 2*n; i++)
		if(!ins[i] && i != piv2)
			nvec.push_back(i);

	// cout << idx << " " << piv1 << " " << piv2 << "\n";
	// print(vec);
	// print(nvec);

	for(int i = 1; i <= 2*n; i++)
	{
		if(i == piv1 || i == piv2) continue;
		if(ins[i])
		{
			vector<int> temp = nvec;
			temp[0] = i;
			if(ask(temp) != bef) ans[i] = ans[piv2];
			else ans[i] = ans[piv2]^1; 
		}
		else
		{
			vector<int> temp = vec;
			temp[idx] = i;
			if(ask(temp) == bef) ans[i] = ans[piv1];
			else ans[i] = ans[piv1]^1; 
		}
	}
	cout << "! ";
	for(int i = 1; i <= 2*n; i++)
	{
		if(ans[i] == 0)
			cout << "R";
		else cout << "B";
	} fflush(stdout);
	// return 0;
}