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;
void init_ios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
const int N = 2 * (int)1e6 + 100;
int n;
pair<int, int> a[N];
vector<int> pkt[N];
int nr(pair<int, int> cord) {
  return 1001 * (cord.second / 1001) + cord.first / 1001;
}
bool cmp(int x, int y) { return a[x] < a[y]; }
bool cmp2(int x, int y) { return a[x] > a[y]; }
int main() {
  init_ios();
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first >> a[i].second;
    int id = nr(a[i]);
    pkt[id].push_back(i);
  }
  for (int i = 0; i < 1001; ++i) {
    if (i % 2 == 0)
      for (int j = 0; j < 1001; ++j) {
        sort(pkt[1001 * i + j].begin(), pkt[1001 * i + j].end(), cmp);
        for (auto it : pkt[1001 * i + j]) cout << it << " ";
      }
    else
      for (int j = 1000; j >= 0; --j) {
        sort(pkt[1001 * i + j].begin(), pkt[1001 * i + j].end(), cmp2);
        for (auto it : pkt[1001 * i + j]) cout << it << " ";
      }
  }
  cout << "\n";
}
