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
#include <bits/stdc++.h>
using namespace std;
struct Punto {
  int x;
  int y;
  int i;
};
bool comparar(Punto p1, Punto p2) {
  int cp1 = p1.y / 1000, cp2 = p2.y / 1000;
  if (cp1 != cp2) return cp1 < cp2;
  if (p1.x == p2.x) return p1.y < p2.y;
  if (cp1 % 2)
    return p1.x < p2.x;
  else
    return p2.x < p1.x;
}
Punto puntos[1100000];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &puntos[i].x, &puntos[i].y);
    puntos[i].i = i + 1;
  }
  sort(puntos, puntos + n, comparar);
  for (int i = 0; i < n; i++) printf("%d ", puntos[i].i);
  printf("\n");
}
