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 <iostream>
#include <vector>

using namespace std;

int n, s, d, m, poz;
string sol(205, '0');
string color;
vector<int> seq, another_seq;

string interrogate(vector<int> &seq) {

    string answer;

    cout << "? ";
    for(int i = 0; i < seq.size(); ++i)
            cout << seq[i] << " ";
    cout << endl;

    cin >> answer;
    return answer;

}

int main() {

    cin >> n;

    for(int i = 1; i <= n; ++i)
            seq.push_back(i);

    color = interrogate(seq);

    seq.clear();

    s = 1;
    d = n + 1;

    while(d - s > 1) {

        m = (s + d) / 2;

        for(int i = m; i < m + n; ++i)
                seq.push_back(i);

        if(interrogate(seq) == color)
                s = m;
        else
                d = m;

        seq.clear();

    }

    poz = s;

    sol[poz] = color[0];
    sol[poz + n] = ((int)'R' + 'B') - color[0];

    for(int i = poz + 1; i < poz + n; ++i)
            seq.push_back(i);

    for(int i = 1; i <= 2 * n; ++i)
            if(i < poz || i > poz + n) {

                seq.push_back(i);
                sol[i] = interrogate(seq)[0];
                seq.pop_back();

                another_seq.push_back(i);

            }

    seq = another_seq;

    for(int i = 1; i <= 2 * n; ++i)
            if(i > poz && i < poz + n) {

                seq.push_back(i);
                sol[i] = interrogate(seq)[0];
                seq.pop_back();

            }

    cout << "! " << string(sol.begin() + 1, sol.begin()+ 2 * n + 1) << endl;
}