#include <bits/stdc++.h>
using namespace std;
bool compare1(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second < b.second.second ||
         (a.second.second == b.second.second &&
          a.second.first < b.second.first);
}
bool compare2(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second > b.second.second ||
         (a.second.second == b.second.second &&
          a.second.first > b.second.first);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, pair<int, int> > >** points =
      new vector<pair<int, pair<int, int> > >*[1001];
  for (int i = 0; i < 1001; i++) {
    points[i] = new vector<pair<int, pair<int, int> > >[1001];
  }
  for (int i = 1; i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    points[a / 1000][b / 1000].push_back(make_pair(i, make_pair(a, b)));
  }
  vector<int> path;
  for (int j = 0; j < 1001; j++) {
    if (j % 2 == 0) {
      for (int i = 0; i < 1001; i++) {
        sort(points[i][j].begin(), points[i][j].end(), compare1);
        for (int k = 0; k < points[i][j].size(); k++) {
          path.push_back(points[i][j][k].first);
        }
      }
    } else {
      for (int i = 1000; i >= 0; i--) {
        sort(points[i][j].begin(), points[i][j].end(), compare2);
        for (int k = 0; k < points[i][j].size(); k++) {
          path.push_back(points[i][j][k].first);
        }
      }
    }
  }
  for (int i = 0; i < path.size(); i++) {
    printf("%d ", path[i]);
  }
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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 compare1(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second < b.second.second ||
         (a.second.second == b.second.second &&
          a.second.first < b.second.first);
}
bool compare2(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second > b.second.second ||
         (a.second.second == b.second.second &&
          a.second.first > b.second.first);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, pair<int, int> > >** points =
      new vector<pair<int, pair<int, int> > >*[1001];
  for (int i = 0; i < 1001; i++) {
    points[i] = new vector<pair<int, pair<int, int> > >[1001];
  }
  for (int i = 1; i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    points[a / 1000][b / 1000].push_back(make_pair(i, make_pair(a, b)));
  }
  vector<int> path;
  for (int j = 0; j < 1001; j++) {
    if (j % 2 == 0) {
      for (int i = 0; i < 1001; i++) {
        sort(points[i][j].begin(), points[i][j].end(), compare1);
        for (int k = 0; k < points[i][j].size(); k++) {
          path.push_back(points[i][j][k].first);
        }
      }
    } else {
      for (int i = 1000; i >= 0; i--) {
        sort(points[i][j].begin(), points[i][j].end(), compare2);
        for (int k = 0; k < points[i][j].size(); k++) {
          path.push_back(points[i][j][k].first);
        }
      }
    }
  }
  for (int i = 0; i < path.size(); i++) {
    printf("%d ", path[i]);
  }
}

```