#include <bits/stdc++.h>
using namespace std;
int n, a, b;
vector<pair<int, int> > x[255][4005];
int main() {
  scanf("%d\n", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d\n", &a, &b);
    x[a / 4000][b / 250].push_back(make_pair(a, i + 1));
  }
  for (int i = 0; i < 252; i++) {
    for (int j = 0; j < 4002; j++)
      if (!x[i][j].empty()) {
        sort(x[i][j].begin(), x[i][j].end());
        for (size_t k = 0; k < x[i][j].size(); k++)
          printf("%d ", x[i][j][k].second);
      }
    i++;
    for (int j = 4001; j >= 0; j--)
      if (!x[i][j].empty()) {
        sort(x[i][j].begin(), x[i][j].end());
        for (size_t k = 0; k < x[i][j].size(); k++)
          printf("%d ", x[i][j][k].second);
      }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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, a, b;
vector<pair<int, int> > x[255][4005];
int main() {
  scanf("%d\n", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d\n", &a, &b);
    x[a / 4000][b / 250].push_back(make_pair(a, i + 1));
  }
  for (int i = 0; i < 252; i++) {
    for (int j = 0; j < 4002; j++)
      if (!x[i][j].empty()) {
        sort(x[i][j].begin(), x[i][j].end());
        for (size_t k = 0; k < x[i][j].size(); k++)
          printf("%d ", x[i][j][k].second);
      }
    i++;
    for (int j = 4001; j >= 0; j--)
      if (!x[i][j].empty()) {
        sort(x[i][j].begin(), x[i][j].end());
        for (size_t k = 0; k < x[i][j].size(); k++)
          printf("%d ", x[i][j][k].second);
      }
  }
  return 0;
}

```