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>
#include <stdlib.h>
#include <math.h>
using namespace std;
typedef long long ll;
const ll mod=1000000007;
const ll BIG=1e15;

int ans[200];
int N; 
int L,R,X;

int ShA(int x){
  cout<<"? "<<flush;
  for(int i=0;i<N;i++) cout<<x+i<<' '<<flush;
  cout<<endl;
  string S; cin>>S;
  if(S=="Red") return 1;
  if(S=="Blue") return -1;
  exit(0);
}

int ShB(int x){
  if(x==L) return X;
  if(x==L+N) return -X;
  cout<<"? "<<flush;
  cout<<x<<' '<<flush;
  if(x<L || L+N<x) for(int i=0;i<N-1;i++) cout<<L+1+i<<' '<<flush;
  else{
    if(L!=1) for(int i=1;i<L;i++) cout<<i<<' '<<flush;
    if(L!=N) for(int i=L+N+1;i<=2*N;i++) cout<<i<<' '<<flush;
  }
  cout<<endl;
  string S; cin>>S;
  if(S=="Red") return 1;
  if(S=="Blue") return -1;
  exit(0);
  
}

int main(){
  cin>>N;
  X=ShA(1);
  L=1, R=N+1;
  while(R>L+1){
    int MID=(R+L)/2;
    if(X==ShA(MID)) L=MID;
    else R=MID;
  }
  //cout<<L<<endl;
  
  for(int i=1;i<=2*N;i++) ans[i]=ShB(i);
  cout<<"! "<<flush;
  for(int i=1;i<=2*N;i++){
    if(ans[i]==1) cout<<'R'<<flush;
    else cout<<'B'<<flush;
  }
  cout<<endl;




}