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;
bool cmp(const pair<pair<int, int>, int> &p,
         const pair<pair<int, int>, int> &q) {
  if (p.first.second != q.first.second) return p.first.second < q.first.second;
  return p.first.first < q.first.first;
}
void Do(vector<pair<pair<int, int>, int> > &a, int l, int r, bool f) {
  if (r - l == 1) {
    printf("%d ", a[l].second);
    return;
  }
  if (f)
    sort(a.begin() + l, a.begin() + r, cmp);
  else
    sort(a.begin() + l, a.begin() + r);
  Do(a, l, (l + r) / 2, !f);
  Do(a, (l + r) / 2, r, !f);
}
int main() {
  vector<pair<pair<int, int>, int> > a;
  int i, n, x, y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d%d", &x, &y);
    a.push_back(make_pair(make_pair(x, y), i + 1));
  }
  Do(a, 0, n, 0);
}
