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 <cstdio>
#include <vector>

using namespace std;

char s[5];
int ans[200];

int ask(vector <int> &v) {
    int i;
    
    putchar('?');
    for (i = 0; i < v.size(); i++) printf(" %d", v[i]);
    puts("");
    fflush(stdout);
    
    scanf("%s", s);
    if (s[0] == 'R') {
        return 0;
    } else {
        return 1;
    }
}

int main() {
    int n, c, x, l, r, m, i, j;
    vector <int> v, w;
    
    scanf("%d", &n);
    
    for (i = 1; i <= n * 2; i++) v.push_back(i);
    
    for (i = 0; i < n; i++) w.push_back(v[i]);
    c = ask(w);
    
    l = 0, r = n, m = (l + r) / 2;
    
    while (r - l > 1) {
        vector <int> z;
        
        for (i = m; i < m + n; i++) z.push_back(v[i]);
        
        if (ask(z) == c) {
            l = m;
            m = (l + r) / 2;
        } else {
            r = m;
            m = (l + r) / 2;
        }
    }
    
    x = r;
    
    ans[x - 1] = c;
    ans[x + n - 1] = 1 - c;
    
    for (i = x; i < x + n - 1; i++) {
        vector <int> z;
        
        for (j = x - 1; j <= x + n - 1; j++) {
            if (j == i) continue;
            z.push_back(v[j]);
        }
        
        if (ask(z) == c) {
            ans[i] = 1 - c;
        } else {
            ans[i] = c;
        }
    }
    
    for (i = 0; i < n * 2; i++) {
        vector <int> z;
        
        if (i >= x - 1 && i <= x + n - 1) continue;
        
        for (j = x; j < x + n - 1; j++) z.push_back(v[j]);
        
        z.push_back(v[i]);
        
        if (ask(z) == c) {
            ans[i] = c;
        } else {
            ans[i] = 1 - c;
        }
    }
    
    printf("! ");
    for (i = 0; i < n * 2; i++) {
        if (ans[i] == 0) {
            putchar('R');
        } else {
            putchar('B');
        }
    }
    puts("");
    fflush(stdout);
    
    return 0;
}