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;
struct Point {
  int x, y, i;
  Point(){};
  Point(int x, int y, int i) : x(x), y(y), i(i){};
};
const int N = (int)1e6 + 10;
const int S = (int)1e3;
int n;
vector<Point> B[N / S + 10];
bool cmp1(Point A, Point B) { return A.y < B.y; }
bool cmp2(Point A, Point B) { return A.y > B.y; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    B[x / S].push_back(Point(x, y, i));
  }
  vector<int> ans;
  for (int k = 0, d = 0; k <= N / S; ++k) {
    sort(B[k].begin(), B[k].end(), (d == 0) ? cmp1 : cmp2);
    d ^= 1;
    for (int i = 0; i < (int)B[k].size(); ++i) ans.push_back(B[k][i].i);
  }
  for (int i = 0; i < (int)ans.size(); ++i) printf("%d ", ans[i]);
  return 0;
}
