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 oo = 1 << 28;
const double err = 1e-8;
struct pto {
  int x, y, p;
};
bool operator<(pto o, pto p) {
  int p1 = o.x / 1000, p2 = p.x / 1000;
  if (p1 != p2) return p1 < p2;
  if (p1 & 1) return o.y < p.y;
  return p.y < o.y;
}
int n;
vector<pto> ptos;
vector<int> up, down;
int dir(pto a, pto b, pto c) {
  return (a.x - b.x) * (a.y * c.y) - (a.x - c.x) * (a.y - b.y);
}
map<pair<int, int>, bool> ma;
int main() {
  scanf("%d", &n);
  int borr = 0;
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    ptos.push_back((pto){x, y, i + 1});
  }
  sort(ptos.begin(), ptos.end());
  for (int i = 0; i < n; i++) printf("%d ", ptos[i].p);
  printf("\n");
  return 0;
}
