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>
const int N = 1000000;
using namespace std;
int x[N], y[N], id[N];
bool cmp(int i, int j) {
  int x1 = x[i] / 1000, x2 = x[j] / 1000;
  return x1 < x2 || (x1 == x2 && (x1 % 2 ? y[i] > y[j] : y[i] < y[j]));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    id[i] = i;
    scanf("%d%d", &x[i], &y[i]);
  }
  sort(id + 1, id + n + 1, cmp);
  for (int i = 1; i <= n; ++i) printf("%d ", id[i]);
  return 0;
}
