#include <bits/stdc++.h>
using namespace std;
struct my {
  int x, y, z, numb;
} a[2000000];
inline char o1(my a, my b) {
  if (a.numb == b.numb && a.numb % 2 == 0) {
    if (a.y > b.y)
      return true;
    else
      return false;
  } else if (a.numb == b.numb && a.numb % 2 == 1) {
    if (a.y < b.y)
      return true;
    else
      return false;
  } else if (a.numb < b.numb)
    return true;
  else
    return false;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    a[i].z = i;
    a[i].numb = a[i].x / 1000;
  }
  sort(a, a + n, o1);
  for (int i = 0; i < n; i++) printf("%d ", a[i].z + 1);
}

### Prompt

Generate 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 my {
  int x, y, z, numb;
} a[2000000];
inline char o1(my a, my b) {
  if (a.numb == b.numb && a.numb % 2 == 0) {
    if (a.y > b.y)
      return true;
    else
      return false;
  } else if (a.numb == b.numb && a.numb % 2 == 1) {
    if (a.y < b.y)
      return true;
    else
      return false;
  } else if (a.numb < b.numb)
    return true;
  else
    return false;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    a[i].z = i;
    a[i].numb = a[i].x / 1000;
  }
  sort(a, a + n, o1);
  for (int i = 0; i < n; i++) printf("%d ", a[i].z + 1);
}

```