#include <bits/stdc++.h>
using namespace std;
bool compare(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  if (a.second.second == b.second.second) {
    return a.second.first < b.second.first;
  }
  return a.second.second < b.second.second;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > bucket[1005];
  int x, y;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    bucket[x / 1000].push_back(make_pair(i, make_pair(x, y)));
  }
  for (int i = 0; i < 1002; i++) {
    sort(bucket[i].begin(), bucket[i].end(), compare);
  }
  for (int i = 0; i < 1002; i++) {
    if (i % 2) {
      for (int j = 0; j < bucket[i].size(); j++) {
        cout << (bucket[i][j].first) + 1 << " ";
      }
    } else {
      for (int j = bucket[i].size() - 1; j >= 0; j--) {
        cout << (bucket[i][j].first) + 1 << " ";
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool compare(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  if (a.second.second == b.second.second) {
    return a.second.first < b.second.first;
  }
  return a.second.second < b.second.second;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > bucket[1005];
  int x, y;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    bucket[x / 1000].push_back(make_pair(i, make_pair(x, y)));
  }
  for (int i = 0; i < 1002; i++) {
    sort(bucket[i].begin(), bucket[i].end(), compare);
  }
  for (int i = 0; i < 1002; i++) {
    if (i % 2) {
      for (int j = 0; j < bucket[i].size(); j++) {
        cout << (bucket[i][j].first) + 1 << " ";
      }
    } else {
      for (int j = bucket[i].size() - 1; j >= 0; j--) {
        cout << (bucket[i][j].first) + 1 << " ";
      }
    }
  }
  return 0;
}

```