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;
//#define cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
	return os << "(" << x.first << "," << x.second << ")";
}
typedef long long ll;
typedef long double ld;
char str[15];
int ask(vector<int>& x) {
	printf("?");
	for (int y : x) printf(" %d", y + 1);
	printf("\n");
	fflush(stdout);
	scanf("%s", str);
	if (str[0] == '-') exit(0);
	return str[0] == 'R';
}
int main() {
	int n;
	scanf("%d", &n);
	vector<int> v(n);
	iota(v.begin(), v.end(), 0);
	int init = ask(v);
	int lo = 0, hi = n - 1;
	while (lo < hi) {
		int m = (lo + hi + 1) >> 1;
		iota(v.begin(), v.end(), m);
		int t = ask(v);
		if (t == init) lo = m;
		else hi = m - 1;
	}
	int m = lo;
	// m+1..m+n-1 have equal R,B.
	iota(v.begin(), v.end(), m + 1);
	vector<int> ans(2 * n, -1);
	for (int i = 0; i < 2 * n; ++i) {
		if (i < m + 1 || i >= m + n) {
			v.back() = i;
			ans[i] = ask(v);
		}
	}
	int c = 0;
	for (int i = 0; i < 2 * n; ++i)
		if (ans[i] == 0 && c < n / 2) v[c++] = i;
	for (int i = 0; i < 2 * n; ++i)
		if (ans[i] == 1 && c < n - 1) v[c++] = i;
	for (int i = m + 1; i < m + n; ++i) {
		v.back() = i;
		ans[i] = ask(v);
	}
	printf("! ");
	for (int i = 0; i < 2 * n; ++i) printf("%c", "BR"[ans[i]]);
	printf("\n");
	fflush(stdout);
}