On a plane are n points (xi, yi) with integer coordinates between 0 and 106. The distance between the two points with numbers a and b is said to be the following value: <image> (the distance calculated by such formula is called Manhattan distance).

We call a hamiltonian path to be some permutation pi of numbers from 1 to n. We say that the length of this path is value <image>.

Find some hamiltonian path with a length of no more than 25 × 108. Note that you do not have to minimize the path length.

Input

The first line contains integer n (1 ≤ n ≤ 106).

The i + 1-th line contains the coordinates of the i-th point: xi and yi (0 ≤ xi, yi ≤ 106).

It is guaranteed that no two points coincide.

Output

Print the permutation of numbers pi from 1 to n — the sought Hamiltonian path. The permutation must meet the inequality <image>.

If there are multiple possible answers, print any of them.

It is guaranteed that the answer exists.

Examples

Input

5
0 7
8 10
3 4
5 0
9 12


Output

4 3 1 2 5 

Note

In the sample test the total distance is:

<image>

(|5 - 3| + |0 - 4|) + (|3 - 0| + |4 - 7|) + (|0 - 8| + |7 - 10|) + (|8 - 9| + |10 - 12|) = 2 + 4 + 3 + 3 + 8 + 3 + 1 + 2 = 26
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class P577E {
    private static class O implements Comparable<O> {
        public final int i;
        public final int q;
        public final int sx;
        public O(int i, int x, int y) {
            this.i = i;
            this.q = y / 1500;
            this.sx = (q % 2 == 0) ? x : -x;
        }

        public int compareTo(O other) {
            int c = Integer.compare(this.q, other.q);
            if (c == 0) {
                c = Integer.compare(this.sx, other.sx);
                if (c == 0) {
                    c = Integer.compare(this.i, other.i);
                }
            }
            return c;
        }
    }

    public static void main(String[] args) throws IOException {
        final long t0 = System.currentTimeMillis();
        final BufferedReader in = new BufferedReader(
            new InputStreamReader(System.in));
        final int n = Integer.parseInt(in.readLine());
        final O[] os = new O[n];
        for (int i = 0; i < n; i++) {
            final String[] ss = in.readLine().split(" ", 2);
            os[i] = new O(i+1,
                          Integer.parseInt(ss[0]),
                          Integer.parseInt(ss[1]));
        }
        Arrays.sort(os);
        final StringBuilder sb = new StringBuilder();
        for (int i = 0; i<n; i++) {
            sb.append(os[i].i);
            sb.append('\n');
        }
        System.out.print(sb);
    }
}
