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>
int n,i,j,l,r,m,lv,rv,nv;
char ans[299],c[9];
int ask(int x){
	printf("? ");
	for(int i=0;i<n;++i)printf("%d ",x-i);puts("");fflush(stdout);
	scanf("%s",c);return *c=='R';
}
char toc(int x){return x==1?'R':'B';}
int main(){
	scanf("%d",&n);
	l=n;r=n*2-1;
	lv=ask(n);rv=ask(n*2);
	for(;l<r;){
		m=l+r+1>>1;
		nv=ask(m);
		if(nv==lv)l=m;else r=m-1;
	}
	ans[l+1]=toc(rv);ans[l-n+1]=toc(lv);
	for(i=1;i<=n*2;++i){
		if(i==l+1 || i==l-n+1)continue;
		if(l-n+1<i && i<=l){
			printf("? ");
			for(j=1;j<=n*2;++j)if(j>l+1 || j<l-n+1)printf("%d ",j);
			printf("%d\n",i);fflush(stdout);
			scanf("%s",c);ans[i]=*c;		
		}else{
			printf("? ");
			for(j=1;j<=n*2;++j)if(l-n+1<j && j<=l)printf("%d ",j);
			printf("%d\n",i);fflush(stdout);
			scanf("%s",c);ans[i]=*c;
		}
	}
	printf("! %s\n",ans+1);fflush(stdout);
	return 0;
}