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>
struct Point {
  long long x;
  long long y;
  int no;
};
int dist(Point a, Point b) {
  return ((a.x - b.x) < 0 ? (-(a.x - b.x)) : (a.x - b.x)) +
         ((a.y - b.y) < 0 ? (-(a.y - b.y)) : (a.y - b.y));
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  std::cin >> n;
  std::vector<Point> x[1001];
  Point p;
  for (int i = 0; i < n; i++) {
    std::cin >> p.x >> p.y;
    p.no = i + 1;
    x[p.x / 1000].push_back(p);
  }
  for (int i = 0; i < 1001; i++) {
    std::sort(x[i].begin(), x[i].end(),
              [](Point a, Point b) { return a.y < b.y; });
  }
  for (int i = 0; i < 1001; i++) {
    if (i % 2) {
      for (auto it = x[i].rbegin(); it != x[i].rend(); ++it) {
        std::cout << (*it).no << " ";
      }
    } else {
      for (auto point : x[i]) std::cout << point.no << " ";
    }
  }
  std::cout << std::endl;
}
