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<iostream>
#include<vector>
#include<algorithm>
#define ep emplace
#define eb emplace_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define semicolon ;
#define ryan bear
using namespace std;
typedef long long ll;
typedef pair<int,int>pi;
typedef pair<ll,ll>pl;
const int inf=1e9+7;
const ll INF=1e18;
int n;
inline bool qu(int i)
{
    cout<<'?';
    for(int j=0;j++<n;)
        cout<<' '<<i+j;
    cout<<endl;
    string s;
    cin>>s;
    return s[0]=='B';
}
inline bool qu(vector<int>&v)
{
    cout<<'?';
    for(int&t:v)
        cout<<' '<<t;
    cout<<endl;
    string s;
    cin>>s;
    return s[0]=='B';
}
int i,j;
int x[1010];
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>n;
    int f=qu(0);
    int s=1;
    int e=n;
    while(s<e)
    {
        int m=s+e>>1;
        if(qu(m)!=f)
            e=m;
        else
            s=m+1;
    }
    for(i=0;i++<n*2;)
        x[i]=-1;
    x[s]=f;
    x[s+n]=!f;
    for(i=0;i++<n*2;)
    {
        if(x[i]!=-1)
            continue;
        if(s<=i&&i<=s+n)
        {
            vector<int>v;
            for(j=s;j<=s+n;j++)
                if(j!=i)
                    v.eb(j);
            int q=qu(v);
            x[i]=!q;
        }
        else
        {
            vector<int>v;
            for(j=s+1;j<s+n;j++)
                v.eb(j);
            v.eb(i);
            int q=qu(v);
            x[i]=q;
        }
    }
    cout<<"! ";
    for(i=0;i++<2*n;)
        cout<<(x[i]==1?'B':'R');
    cout<<endl;
    return 0;
}