#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
struct Point {
  int x, y;
  int ind;
};
bool cmp(const Point& p1, const Point& p2) {
  if (p1.x / 1000 != p2.x / 1000) return p1.x / 1000 < p2.x / 1000;
  if (p1.x / 1000 % 2 == 0) {
    if (p1.y != p2.y) return p1.y < p2.y;
    return p1.x < p2.x;
  } else {
    if (p1.y != p2.y) return p1.y > p2.y;
    return p1.x < p2.x;
  }
  return false;
}
Point points[1000010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    points[i].ind = i + 1;
    scanf("%d%d", &points[i].x, &points[i].y);
  }
  sort(points, points + n, cmp);
  for (int i = 0; i < n; i++) printf("%d ", points[i].ind);
  return 0;
}

### Prompt

Develop a solution in CPP 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>
#pragma warning(disable : 4996)
using namespace std;
struct Point {
  int x, y;
  int ind;
};
bool cmp(const Point& p1, const Point& p2) {
  if (p1.x / 1000 != p2.x / 1000) return p1.x / 1000 < p2.x / 1000;
  if (p1.x / 1000 % 2 == 0) {
    if (p1.y != p2.y) return p1.y < p2.y;
    return p1.x < p2.x;
  } else {
    if (p1.y != p2.y) return p1.y > p2.y;
    return p1.x < p2.x;
  }
  return false;
}
Point points[1000010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    points[i].ind = i + 1;
    scanf("%d%d", &points[i].x, &points[i].y);
  }
  sort(points, points + n, cmp);
  for (int i = 0; i < n; i++) printf("%d ", points[i].ind);
  return 0;
}

```