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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
const int maxn = 1000000 + 100;
int x[maxn];
int y[maxn];
vector<pair<int, int> > b[maxn / 1000 + 50];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    scanf("%d%d", &x[i], &y[i]);
    int f = x[i] / 1000;
    if (f % 2 == 0)
      b[f].push_back(make_pair(y[i], i + 1));
    else
      b[f].push_back(make_pair(-y[i], i + 1));
  }
  for (int f = 0; f < (1010); f++) {
    sort((b[f]).begin(), (b[f]).end());
    for (int i = 0; i < (((int)(b[f]).size())); i++)
      cout << b[f][i].second << " ";
  }
  cout << '\n';
  return 0;
}
