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;
  int y;
  int num;
};
bool cmp(Point x1, Point x2) { return x1.y < x2.y; }
int main() {
  int n;
  scanf("%d", &n);
  array<vector<struct Point>, 1001> groups;
  struct Point q;
  int x, y;
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &x, &y);
    q.x = x;
    q.y = y;
    q.num = i + 1;
    groups[x / 1000].push_back(q);
  }
  for (int i = 0; i < 1001; i++) {
    sort(groups[i].begin(), groups[i].end(), cmp);
  }
  bool reverse = false;
  for (int i = 0; i < 1001; i++) {
    if (reverse == false) {
      for (auto it = groups[i].begin(); it != groups[i].end(); ++it) {
        printf("%d ", (*it).num);
      }
      reverse = true;
    } else {
      for (auto it = groups[i].rbegin(); it != groups[i].rend(); ++it) {
        printf("%d ", (*it).num);
      }
      reverse = false;
    }
  }
  printf("\n");
  return 0;
}
