#include <bits/stdc++.h>
using namespace std;
int n;
int x[1000000], y[1000000];
vector<pair<pair<int, int>, int> > v;
vector<pair<int, int> > s[1001];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &x[i], &y[i]);
    v.push_back(make_pair(make_pair(x[i], y[i]), i + 1));
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    s[(v[i].first.first - 1) / 1000].push_back(
        make_pair(v[i].first.second, v[i].second));
  }
  int l = 0;
  for (int i = 0; i < 1000; i++) {
    if ((int)s[i].size() > 0) {
      l++;
      sort(s[i].begin(), s[i].end());
      if (l % 2 == 1) {
        for (int j = 0; j < (int)s[i].size(); j++) {
          printf("%d ", s[i][j].second);
        }
      } else {
        for (int j = (int)s[i].size() - 1; j >= 0; j--) {
          printf("%d ", s[i][j].second);
        }
      }
    }
  }
  cout << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 n;
int x[1000000], y[1000000];
vector<pair<pair<int, int>, int> > v;
vector<pair<int, int> > s[1001];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &x[i], &y[i]);
    v.push_back(make_pair(make_pair(x[i], y[i]), i + 1));
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    s[(v[i].first.first - 1) / 1000].push_back(
        make_pair(v[i].first.second, v[i].second));
  }
  int l = 0;
  for (int i = 0; i < 1000; i++) {
    if ((int)s[i].size() > 0) {
      l++;
      sort(s[i].begin(), s[i].end());
      if (l % 2 == 1) {
        for (int j = 0; j < (int)s[i].size(); j++) {
          printf("%d ", s[i][j].second);
        }
      } else {
        for (int j = (int)s[i].size() - 1; j >= 0; j--) {
          printf("%d ", s[i][j].second);
        }
      }
    }
  }
  cout << endl;
  return 0;
}

```