#include <bits/stdc++.h>
using namespace std;
bool fct(const pair<int, pair<int, int> >& p1,
         const pair<int, pair<int, int> >& p2) {
  if (p1.first / 1000 == p2.first / 1000) {
    if ((p1.first / 1000) % 2 == 0)
      return p1.second.first < p2.second.first;
    else
      return p1.second.first > p2.second.first;
  }
  return p1.first / 1000 < p2.first / 1000;
}
int dist(pair<int, int> p1, pair<int, int> p2) {
  return abs(p1.first - p2.first) + abs(p1.second - p2.second);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, pair<int, int> > > v;
  for (int i(1); i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    v.push_back(make_pair(a, make_pair(b, i)));
  }
  sort(v.begin(), v.end(), fct);
  for (int i(0); i < n; i++) printf("%d ", (v[i].second).second);
  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;
bool fct(const pair<int, pair<int, int> >& p1,
         const pair<int, pair<int, int> >& p2) {
  if (p1.first / 1000 == p2.first / 1000) {
    if ((p1.first / 1000) % 2 == 0)
      return p1.second.first < p2.second.first;
    else
      return p1.second.first > p2.second.first;
  }
  return p1.first / 1000 < p2.first / 1000;
}
int dist(pair<int, int> p1, pair<int, int> p2) {
  return abs(p1.first - p2.first) + abs(p1.second - p2.second);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, pair<int, int> > > v;
  for (int i(1); i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    v.push_back(make_pair(a, make_pair(b, i)));
  }
  sort(v.begin(), v.end(), fct);
  for (int i(0); i < n; i++) printf("%d ", (v[i].second).second);
  return 0;
}

```