#include <bits/stdc++.h>
using namespace std;
struct pct {
  long long x, y, id;
  pct(int x, int y, int id) {
    this->x = x;
    this->y = y;
    this->id = id;
  }
};
long long n, i, x, y;
vector<pct> v;
bool comp(pct a, pct b) {
  if (a.x / 1000 != b.x / 1000) return a.x < b.x;
  if ((a.x / 1000) % 2)
    return a.y < b.y;
  else
    return a.y > b.y;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x >> y;
    v.push_back(pct(x, y, i));
  }
  sort(v.begin(), v.end(), comp);
  for (auto it : v) cout << it.id << ' ';
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
struct pct {
  long long x, y, id;
  pct(int x, int y, int id) {
    this->x = x;
    this->y = y;
    this->id = id;
  }
};
long long n, i, x, y;
vector<pct> v;
bool comp(pct a, pct b) {
  if (a.x / 1000 != b.x / 1000) return a.x < b.x;
  if ((a.x / 1000) % 2)
    return a.y < b.y;
  else
    return a.y > b.y;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x >> y;
    v.push_back(pct(x, y, i));
  }
  sort(v.begin(), v.end(), comp);
  for (auto it : v) cout << it.id << ' ';
  return 0;
}

```