#include <bits/stdc++.h>
using namespace std;
int BLOCK = 1000;
struct Point {
  int x, y, idx;
};
int N;
Point pts[1000000 + 1];
bool cmp(Point a, Point b) {
  if (a.x / BLOCK != b.x / BLOCK) return a.x / BLOCK < b.x / BLOCK;
  return (a.x / BLOCK & 1) ? (a.y < b.y) : (a.y > b.y);
}
int main() {
  cin.tie(0)->sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> pts[i].x >> pts[i].y, pts[i].idx = i;
  sort(pts + 1, pts + 1 + N, cmp);
  for (int i = 1; i <= N; i++) cout << pts[i].idx << " ";
  cout << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int BLOCK = 1000;
struct Point {
  int x, y, idx;
};
int N;
Point pts[1000000 + 1];
bool cmp(Point a, Point b) {
  if (a.x / BLOCK != b.x / BLOCK) return a.x / BLOCK < b.x / BLOCK;
  return (a.x / BLOCK & 1) ? (a.y < b.y) : (a.y > b.y);
}
int main() {
  cin.tie(0)->sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> pts[i].x >> pts[i].y, pts[i].idx = i;
  sort(pts + 1, pts + 1 + N, cmp);
  for (int i = 1; i <= N; i++) cout << pts[i].idx << " ";
  cout << "\n";
  return 0;
}

```