#include <fstream>
#include <vector>
#include <algorithm>
using namespace std;


class Task {
 public:
	void solve() {
		read_input();
		print_output(get_result());
	}

 private:
	int n;
  vector<int> mysol;
	void read_input() {
		ifstream fin("in");
		fin >> n;
		fin.close();
	}
  int myabs(int x) {
    if(x < 0) {
      return -x;
    }
    return x;
  }

  int check(std::vector<int> sol, int k) {
      for(int i = 1; i < k; i++) {
        if(sol[k] == sol[i]) {
          return 0;
        }
        if((k - i) == myabs(sol[k] - sol[i])) {
          return 0;
        }
      }
      return 1;
  }

   void bkt(std::vector<int> &sol, int k, int &ok) {
     if(k == n + 1) {
       if(ok == 0) {
         mysol = sol;
         ok = 1;
       }
     } else {
       for(int i = 1; i <= n; i++) {
         sol[k] = i;
         if(check(sol, k) == 1) {
           bkt(sol, k + 1, ok);
         }
       }
     }
   }

	vector<int> get_result() {
		vector<int> sol(n + 1, 0);
    mysol.resize(n + 1);
    int ok = 0;
    bkt(sol, 1, ok);

		/*
		TODO: Gasiti o solutie pentru problema damelor pe o tabla de dimensiune
		n x n.

		Pentru a plasa o dama pe randul i, coloana j:
			sol[i] = j.
		Randurile si coloanele se indexeaza de la 1 la n.

		De exemplu, configuratia (n = 5)
		X----
		--X--
		----X
		-X---
		---X-
		se va reprezenta prin sol[1..5] = {1, 3, 5, 2, 4}.
		*/

		return mysol;
	}

	void print_output(vector<int> result) {
		ofstream fout("out");
		for (int i = 1; i <= n; i++) {
			fout << result[i] << (i != n ? ' ' : '\n');
		}
		fout.close();
	}
};

int main() {
	Task task;
	task.solve();
	return 0;
}