#include <bits/stdc++.h>
using namespace std;
struct Point {
  int id, x, y;
};
vector<Point> block[1010];
int cmp(Point a, Point b) { return a.y < b.y; }
int main(int n) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    Point t;
    scanf("%d%d", &t.x, &t.y);
    t.id = i;
    block[t.x / 1000].push_back(t);
  }
  bool flag = 0;
  for (int i = 0; i <= 1000; i++) {
    sort(block[i].begin(), block[i].end(), cmp);
    if (block[i].size()) {
      int len = block[i].size();
      if (!flag)
        for (int j = 0; j < len; j++) printf("%d ", block[i][j].id);
      else
        for (int j = len - 1; j > -1; j--) printf("%d ", block[i][j].id);
      flag ^= 1;
    }
  }
  return 0;
}

### Prompt

Please formulate a cpp solution 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;
struct Point {
  int id, x, y;
};
vector<Point> block[1010];
int cmp(Point a, Point b) { return a.y < b.y; }
int main(int n) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    Point t;
    scanf("%d%d", &t.x, &t.y);
    t.id = i;
    block[t.x / 1000].push_back(t);
  }
  bool flag = 0;
  for (int i = 0; i <= 1000; i++) {
    sort(block[i].begin(), block[i].end(), cmp);
    if (block[i].size()) {
      int len = block[i].size();
      if (!flag)
        for (int j = 0; j < len; j++) printf("%d ", block[i][j].id);
      else
        for (int j = len - 1; j > -1; j--) printf("%d ", block[i][j].id);
      flag ^= 1;
    }
  }
  return 0;
}

```