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;

typedef long long LL;

const int N=99;

int n;

void into(){
  scanf("%d",&n);
}

int ans[N*2+9];

int Query(int p){
  cout<<"? ";
  for (int i=0;i<n;++i) cout<<p+i<<' ';
  cout<<endl;
  fflush(stdout);
  string s;
  cin>>s;
  return s[0]=='R';
}

int Query(int l,int r,int p0){
  cout<<"? ";
  if (l<p0&&p0<r){
	for (int i=l;i<p0;++i) cout<<i<<' ';
	for (int i=p0+1;i<=r;++i) cout<<i<<' ';
	cout<<endl;
  }else{
	for (int i=l+1;i<r;++i) cout<<i<<' ';
	cout<<p0<<endl;
  }
  fflush(stdout);
  string s;
  cin>>s;
  return s[0]=='R'^(l<p0&&p0<r);
}

void Two_divide_ans(){
  int l=1,r=n+1,tl=Query(l),tr=Query(r);
  for (int mid,tm;r-l>1;mid=l+r>>1,(tm=Query(mid))==tl?l=mid:r=mid);
  ans[l]=tl;ans[l+n]=tr;
  for (int i=1;i<=n<<1;++i)
	if (i^l&&i^l+n) ans[i]=Query(l,l+n,i);
}

void work(){
  Two_divide_ans();
}

void outo(){
  cout<<"! ";
  for (int i=1;i<=n<<1;++i)
	cout<<(ans[i]?'R':'B');
  cout<<endl;
}

int main(){
  into();
  work();
  outo();
  return 0;
}