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;
const int N=205;
int n,a[N],b[N];
vector<int>p;
char s[10];
int R()
{
    fflush(stdout);
    scanf("%s",s);
    if(s[0]=='R') return 1;
    return 2;
}
int ask()
{
    printf("?");
    for(int i=0;i<p.size();i++)
        printf(" %d",p[i]);
    printf("\n");
    return R();
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        p.push_back(i);
    int t=ask();
    if(n==1)
    {
        if(t==1) printf("! RB\n");
        else printf("! BR\n");
        return 0;
    }
    int l=1,r=n+1;
    while(r-l>1)
    {
        int m=l+r>>1;
        p.clear();
        for(int i=m;i<=m+n-1;i++)
            p.push_back(i);
        if(ask()==t) l=m;
        else r=m;
    }
    p.clear();
    for(int i=l+1;i<=l+n-1;i++)
        p.push_back(i);
    for(int i=1;i<=2*n;i++)
        if(b[i]==0&&!(i>=l+1&&i<=l+n-1))
    {
        p.push_back(i);
        b[i]=ask();
        p.pop_back();
    }
    p.clear();
    int s1=n/2,s2=n/2;
    for(int i=1;i<=2*n;i++)
        if(b[i]==1&&s1)
    {
        s1--;p.push_back(i);
    }
    else if(b[i]==2&&s2)
    {
        s2--;p.push_back(i);
    }
    for(int i=1;i<=2*n;i++)
        if(!b[i])
        {
            p.push_back(i);
            b[i]=ask();
            p.pop_back();
        }
    printf("! ");
    for(int i=1;i<=2*n;i++)
        if(b[i]==1) putchar('R');
    else putchar('B');
    putchar('\n');
}