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 pt {
  int x, y, id;
  bool operator<(const pt &a) const { return x < a.x; }
};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  const int N = 1e6;
  vector<vector<pt>> a(N);
  for (int i = 0, x, y; i < n; ++i) {
    cin >> x >> y;
    int px = min(999, x / 1000);
    int py = min(999, y / 1000);
    a[px * 1000 + py].push_back({x, y, i + 1});
  }
  vector<int> ans;
  for (int i = 0; i < 1000; ++i) {
    vector<int> tans;
    for (int j = 0; j < 1000; ++j) {
      auto &t = a[i * 1000 + j];
      sort(t.begin(), t.end());
      for (auto k : t) tans.push_back(k.id);
    }
    if (i % 2 == 1) reverse(tans.begin(), tans.end());
    for (auto j : tans) ans.push_back(j);
  }
  for (auto &i : ans) cout << i << " \n"[&i == &ans.back()];
  return 0;
}
