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;
using ll = long long;
void setIO() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
}
int B;
bool cmp(const tuple<int, int, int>& a, const tuple<int, int, int>& b) {
  int id1, l1, r1;
  tie(id1, l1, r1) = a;
  int id2, l2, r2;
  tie(id2, l2, r2) = b;
  if (l1 / B != l2 / B)
    return l1 / B < l2 / B;
  else if ((l1 / B) % 2)
    return r1 > r2;
  else
    return r1 < r2;
}
int main() {
  setIO();
  int n;
  cin >> n;
  int mx = 0;
  vector<tuple<int, int, int>> points(n);
  for (int i = (0); i < (n); i++) {
    int x, y;
    cin >> x >> y;
    points[i] = make_tuple(i, x, y);
    mx = max(mx, x);
  }
  B = sqrt(mx + 1);
  sort((points).begin(), (points).end(), cmp);
  for (int i = (0); i < (n); i++) {
    cout << get<0>(points[i]) + 1 << ' ';
  }
  cout << '\n';
  return 0;
}
