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
#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x, y, i;
  Point(){};
  Point(int xx, int yy, int ii) {
    x = xx;
    y = yy;
    i = ii;
  }
} p[1000005];
bool cmp(Point a, Point b) {
  if (a.x / 1000 == b.x / 1000) {
    if (a.x / 1000 % 2 == 0) {
      if (a.y < b.y) return true;
      return false;
    } else {
      if (a.y > b.y) return true;
      return false;
    }
  } else {
    if (a.x < b.x) return true;
    return false;
  }
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &p[i].x, &p[i].y);
      p[i].i = i + 1;
    }
    sort(p, p + n, cmp);
    printf("%d", p[0].i);
    for (int i = 1; i < n; i++) printf(" %d", p[i].i);
    puts("");
  }
  return 0;
}
