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;
vector<bool> Primes(int n) {
  vector<bool> ans(n + 1, true);
  ans[1] = false;
  for (int i = 2; i * i <= n; i++)
    if (ans[i])
      for (int j = i * i; j <= n; j += i) ans[j] = false;
  return ans;
}
struct point {
  int x;
  int y;
  long long kof1;
  long long kof2;
  int num;
};
bool compare(point a, point b) {
  return (a.kof1 < b.kof1 || (a.kof1 == b.kof1 && a.kof2 < b.kof2));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<point> v(n);
  for (int i = 0; i < n; ++i) {
    long long a, b;
    cin >> a >> b;
    v[i].num = i + 1;
    v[i].kof1 = a / 1001;
    if (a / 1000 % 2 == 0) {
      v[i].kof2 = b;
    } else {
      v[i].kof2 = -b;
    }
  }
  sort(v.begin(), v.end(), compare);
  for (int i = 0; i < n; ++i) {
    printf("%d ", v[i].num);
  }
}
