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 pi 3.1415926535897932384626
using namespace std;
      
struct P{
  long long x,y;
  bool operator < (const P &a )const{
    //if(y!=a.y)
    return y>a.y;
  };
};

int a,b,c,d,i,k,n,m,e,dx[10]={1,0,-1,0,1,1,-1,-1},dy[10]={0,1,0,-1,1,-1,1,-1};
int o[301];
int l[10];
int j[302];

long long x,y,z,mod=1000000007;
//string r;
char r[111];

P u[1];
queue<int> q;

//stack<int> s;
set<int> s;
map<int,int> p;
vector<P> v[200101];

bool as(P a,P b)
{
  if(a.x==b.x)
  return a.y<b.y;
  return a.x<b.x;
}

int f()
{
  printf("? ");
  for(int t=1;t<=a;t++)
    printf("%d ",j[t]);
  puts("");
  fflush(stdout);
  
  
  
  
  scanf("%s",r);
  if(r[0]=='R') return 1;
  return 0;
  
}

int main()
{
  scanf("%d",&a);
  iota(j+1,j+a+1,1);
  b=f();
  n=1,m=a;
  for(;n<m;)
  {
    int h=(n+m)/2+1;
    iota(j+1,j+a+1,h);
    c=f();
    if(c==b) n=h;
    else m=h-1;
  }
  memset(o,-1,sizeof(o));
  o[n]=b;
  o[n+a]=1-b;
  for(int t=1;t<=2*a;t++)
  if(o[t]<0)
  {
    if(t<n||t>n+a)
    {
      iota(j+1,j+a,n+1);
      j[a]=t;
      o[t]=f();
    }
    else
    {
      for(int t=1,w=1,h=a+n+1;t<=a;t++)
      {
        if(w<n) j[t]=w,w++;
        else j[t]=h,h++;
      }
      j[a]=t;
      o[t]=f();
    }
  }
  printf("! ");
  for(int t=1;t<=2*a;t++)
  if(o[t]==1) printf("R");
  else if(o[t]==0) printf("B");
  else printf("#");
  puts("");
  fflush(stdout);
}