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 lig, col;
  Point(int _lig = 0, int _col = 0) {
    lig = _lig;
    col = _col;
  }
};
vector<int> colonnes[1001];
vector<Point> points;
bool comp(const int &a, const int &b) {
  if (points[a].lig == points[b].lig) return points[a].col < points[b].col;
  return points[a].lig < points[b].lig;
}
int main() {
  int nbPoints;
  scanf("%d", &nbPoints);
  int lig, col;
  for (int iPoint = 0; iPoint < nbPoints; iPoint++) {
    scanf("%d%d", &lig, &col);
    points.push_back(Point(lig, col));
    colonnes[col / 1000].push_back(iPoint);
  }
  for (int iColonne = 0; iColonne <= 1000; iColonne++) {
    sort(colonnes[iColonne].begin(), colonnes[iColonne].end(), comp);
    if ((iColonne & 1) == 0)
      for (int iPoint = 0; iPoint < (int)colonnes[iColonne].size(); iPoint++)
        printf("%d ", colonnes[iColonne][iPoint] + 1);
    else
      for (int iPoint = (int)colonnes[iColonne].size() - 1; iPoint >= 0;
           iPoint--)
        printf("%d ", colonnes[iColonne][iPoint] + 1);
  }
  return 0;
}
