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 long long int MINL = (long long int)25 * 10000 * 10001;
const int MAXN = 1000 * 1001;
struct Point {
  int i, x, y;
};
bool lowerP(const Point& p1, const Point& p2) { return p1.y < p2.y; }
bool greaterP(const Point& p1, const Point& p2) { return p1.y > p2.y; }
int n;
vector<int> ans;
vector<Point> lay[1001];
inline long long int dist(const Point& p1, const Point& p2) {
  return abs(p1.x - p2.x) + abs(p1.y - p2.y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int a = 0; a < n; ++a) {
    int x, y;
    cin >> x >> y;
    lay[x / 1000].push_back({a, x, y});
  }
  for (int a = 0; a < 1001; ++a) {
    if (a & 1)
      sort(lay[a].begin(), lay[a].end(), lowerP);
    else
      sort(lay[a].begin(), lay[a].end(), greaterP);
    for (auto b : lay[a]) {
      ans.push_back(b.i);
    }
  }
  for (auto a : ans) {
    cout << a + 1 << " ";
  }
  return 0;
}
