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 = 400005;
int n;
struct point {
  int id, x, y;
  bool operator<(point a) const { return y < a.y; }
};
vector<point> qe[1005];
int main() {
  cin >> n;
  int i, j, k1, k2;
  int fg1[1005];
  memset(fg1, 0, sizeof(fg1));
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &k1, &k2);
    int fg = k1 / 1000 + 1;
    point st;
    st.x = k1, st.y = k2;
    st.id = i;
    qe[fg].push_back(st);
  }
  for (i = 1; i <= 1001; i++) {
    if (!qe[i].empty()) {
      sort(qe[i].begin(), qe[i].end());
      if (i % 2 == 0) {
        for (j = 0; j < qe[i].size(); j++) {
          printf("%d ", qe[i][j].id);
        }
      } else {
        for (j = qe[i].size() - 1; j >= 0; j--) {
          printf("%d ", qe[i][j].id);
        }
      }
    }
  }
  printf("\n");
  return 0;
}
