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 int N = 1e6 + 5;
std::vector<int> v[N];
pair<int, int> a[N];
int E = 0;
bool cmp(int x, int y) {
  int temp = a[x].first / 1050;
  if (temp % 2) {
    return a[x].second < a[y].second;
  }
  return a[x].second > a[y].second;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].first);
    ;
    scanf("%d", &a[i].second);
    ;
    v[a[i].first / 1050].push_back(i);
  }
  std::vector<int> ans;
  for (int i = 0; i <= 1000000; i++) {
    sort(v[i].begin(), v[i].end(), cmp);
    for (auto x : v[i]) {
      ans.push_back(x);
    }
  }
  for (auto x : ans) {
    printf("%d ", x + 1);
  }
  printf("\n");
}
