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;
int n;
int x[1000111], y[1000111];
vector<int> blk[1001];
bool compLess(const int &le, const int &ri) { return (y[le] < y[ri]); }
bool compMore(const int &le, const int &ri) { return (y[le] > y[ri]); }
bool input() {
  if (cin >> n) {
    for (int i = (int)(0), _t = (int)(n - 1); i <= _t; i++) cin >> x[i] >> y[i];
    return true;
  } else {
    return false;
  }
}
void solve() {
  for (int i = (int)(0), _t = (int)(1001 - 1); i <= _t; i++) blk[i].clear();
  for (int i = (int)(0), _t = (int)(n - 1); i <= _t; i++)
    blk[x[i] / 1000].push_back(i);
  for (int i = (int)(0), _t = (int)(1001 - 1); i <= _t; i++) {
    sort((blk[i]).begin(), (blk[i]).end(), (i & 1) ? compMore : compLess);
    for (int j = (int)(0), _t = (int)(((int)((blk[i]).size())) - 1); j <= _t;
         j++)
      cout << (blk[i][j] + 1) << " ";
  }
  cout << endl;
}
int main() {
  ios ::sync_with_stdio(false);
  while (input()) solve();
  return 0;
}
