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 ROW_SIZE = 2501;
const int MAX_N = 1e6 + 10;
struct Point {
  int x, y;
  int ind;
  static bool cmp_even(Point a, Point b) { return a.x < b.x; }
  static bool cmp_odd(Point a, Point b) { return a.x > b.x; }
};
Point points[MAX_N];
vector<Point> buckets[ROW_SIZE];
int pointCnt;
int main() {
  scanf("%i", &pointCnt);
  int row;
  for (int i = 0; i < pointCnt; i++) {
    scanf("%i %i", &points[i].x, &points[i].y);
    points[i].ind = i + 1;
    row = points[i].y / ROW_SIZE;
    buckets[row].push_back(points[i]);
  }
  for (int i = 0; i < ROW_SIZE; i++) {
    if (buckets[i].size() > 1) {
      if (i & 1) {
        sort(buckets[i].begin(), buckets[i].end(), Point::cmp_odd);
      } else {
        sort(buckets[i].begin(), buckets[i].end(), Point::cmp_even);
      }
    }
  }
  for (int i = 0; i < ROW_SIZE; i++) {
    for (Point p : buckets[i]) {
      printf("%i ", p.ind);
    }
  }
}
