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;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
typedef struct $ {
  int first;
  int second;
  int index;
} pp;
bool cmpfun(pp x, pp y) {
  int k = 1009;
  if ((x.first) / k == (y.first) / k) return (x.second < y.second);
  return (x.first < y.first);
}
int main() {
  int t;
  scanf("%d", &t);
  pp p1;
  pp *p = (pp *)malloc((t + 1) * sizeof(pp));
  for (int i = 0; i < t; i++) {
    scanf("%d %d", &p[i].first, &p[i].second);
    p[i].index = i + 1;
  }
  sort(p, p + t, cmpfun);
  for (int i = 0; i < t; i++) printf("%d ", p[i].index);
  return 0;
}
