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 gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int abs(int a) {
  if (a < 0) return -a;
  return a;
}
int i;
bool ord(pair<int, int> a, pair<int, int> b) {
  if (i % 2 == 0) return a.first < b.first;
  return a.first > b.first;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > p[1003];
  int x, y;
  for (i = 0; i < n; i++) {
    scanf("%d%d", &x, &y);
    p[x / 1000].push_back(make_pair(y, i + 1));
  }
  for (i = 0; i < 1003; i++) sort(p[i].begin(), p[i].end(), ord);
  for (i = 0; i < 1003; i++)
    for (int j = 0; j < p[i].size(); j++) cout << p[i][j].second << ' ';
}
