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;
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3fLL;
pair<int, int> pts[1000005];
vector<int> buckets[1005];
bool cmp_up(int a, int b) {
  if (pts[a].second != pts[b].second) return pts[a].second < pts[b].second;
  return pts[a].first < pts[b].first;
}
bool cmp_down(int a, int b) {
  if (pts[a].second != pts[b].second) return pts[a].second > pts[b].second;
  return pts[a].first < pts[b].first;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d %d", &pts[i].first, &pts[i].second);
    buckets[pts[i].first / 1000].push_back(i);
  }
  for (int i = (0); i < (1005); ++i)
    sort(buckets[i].begin(), buckets[i].end(), (i & 1) ? cmp_down : cmp_up);
  bool first = true;
  for (int i = (0); i < (1005); ++i) {
    for (int j = (0); j < (buckets[i].size()); ++j) {
      if (!first) putchar(' ');
      first = false;
      printf("%d", buckets[i][j] + 1);
    }
  }
  puts("");
  return 0;
}
