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 MAXN = 205;

int N;
char s[5], res[MAXN];

int main(){
	scanf("%d", &N);
	
	int prv, cur;
	
	printf("?");
	for(int i = 1; i <= N; i++){
		printf(" %d", i);
	}
	printf("\n");
	fflush(stdout);
	
	scanf("%s", s);
	prv = (s[0] == 'R');
	
	int left = 1, right = N;
	while(left < right){
		int mid = (left + right + 1) / 2;
		
		printf("?");
		for(int j = 1; j <= N; j++){
			printf(" %d", mid + j - 1);
		}
		printf("\n");
		fflush(stdout);
		
		scanf("%s", s);
		cur = (s[0] == 'R');
		
		if(cur == prv){
			left = mid;
		}else{
			right = mid - 1;
		}
	}
	int L = left + 1, R = left + N - 1;
	
	if(L > 1)res[L - 1] = prv ? 'R' : 'B';
	if(R < 2 * N)res[R + 1] = prv ? 'B' : 'R';
	
	for(int i = 1; i < L - 1; i++){
		printf("? %d", i);
		for(int j = L; j <= R; j++){
			printf(" %d", j);
		}
		printf("\n");
		fflush(stdout);
		
		scanf("%s", s);
		cur = (s[0] == 'R');
	
		res[i] = cur ? 'R' : 'B';
	}
	for(int i = R + 2; i <= 2 * N; i++){
		printf("? %d", i);
		for(int j = L; j <= R; j++){
			printf(" %d", j);
		}
		printf("\n");
		fflush(stdout);
		
		scanf("%s", s);
		cur = (s[0] == 'R');
	
		res[i] = cur ? 'R' : 'B';
	}
	
	for(int i = L; i <= R; i++){
		printf("? %d", i);
		
		for(int j = 1; j < L - 1; j++){
			printf(" %d", j);
		}
		for(int j = R + 2; j <= 2 * N; j++){
			printf(" %d", j);
		}
		printf("\n");
		fflush(stdout);
		
		scanf("%s", s);
		cur = (s[0] == 'R');

		res[i] = cur ? 'R' : 'B';
	}
	
	res[2 * N + 1] = 0;
	printf("! %s\n", res + 1);
	
	return 0;
}