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;
typedef long long li;


li n;

bool isred(const vector<li>& query) {
    cout << "?";
    for (int i = 0; i < n; ++i) {
        cout << " " << query[i] + 1;
    }
    cout << endl;
    string response;
    cin >> response;
    return response[0] == 'R';
}

int main() {
    cin >> n;

    vector<li> first_query;
    for (int i = 0; i < n; ++i) {
        first_query.push_back(i);
    }

    bool first_response = isred(first_query);

    li l = 0, r = n;
    while (r - l > 1) {
        li m = (l + r) / 2;
        vector<li> query;
        for (int i = 0; i < n; ++i) {
            query.push_back(m + i);
        }
        bool response = isred(query);
        if (first_response == response) {
            l = m;
        } else {
            r = m;
        }
    }

    vector<bool> ans(2 * n);
    for (int i = 0; i < 2 * n; ++i) {
        vector<li> query;
        bool mask = false;
        if (l + 1 <= i && i < l + n) {
            mask = true;
            for (int j = l; j <= l + n; ++j) {
                if (j != i) {
                    query.push_back(j);
                }
            }
        } else {
            mask = false;
            for (int j = l + 1; j < l + n; ++j) {
                query.push_back(j);
            }
            query.push_back(i);
        }
        sort(query.begin(), query.end());
        ans[i] = isred(query) != mask;

    }

    cout << "! ";
    for (int i = 0; i < 2 * n; ++i) {
        if (ans[i]) {
            cout << "R";
        } else {
            cout << "B";
        }
    }
    cout << endl;
    return 0;
}