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;
vector<long long> c;
vector<vector<long long> > b(1002, c);
vector<vector<vector<long long> > > a(1002, b);
vector<bool> d(1002, false);
vector<vector<bool> > used(1002, d);
long long n;
void print(long long x, long long y) {
  long long i;
  used[x][y] = true;
  for (i = 0; i < a[x][y].size(); i++) cout << a[x][y][i] << " ";
}
pair<long long, long long> nearest(long long x, long long y) {
  pair<long long, long long> ans;
  ans.first = -1;
  ans.second = -1;
  for (long long d = 1; d <= 2000; d++) {
    for (long long dx = -d; dx <= d; dx++) {
      long long dy = d - abs(dx);
      long long newx = x + dx;
      long long newy = y + dy;
      if (newx >= 0 && newy >= 0 && newx <= 1000 && newy <= 1000 &&
          !used[newx][newy])
        return make_pair(newx, newy);
    }
    for (long long dx = -d; dx <= d; dx++) {
      long long dy = -(d - abs(dx));
      long long newx = x + dx;
      long long newy = y + dy;
      if (newx >= 0 && newy >= 0 && newx <= 1000 && newy <= 1000 &&
          !used[newx][newy])
        return make_pair(newx, newy);
    }
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  long long i, j;
  cin >> n;
  for (i = 1; i <= n; i++) {
    long long q, w;
    cin >> q >> w;
    long long row = q / 1000;
    long long col = w / 1000;
    a[row][col].push_back(i);
  }
  for (i = 0; i <= 1000; i++) {
    for (j = 0; j <= 1000; j++)
      if (a[i][j].size() == 0) used[i][j] = true;
  }
  long long curx = -1, cury = -1;
  for (i = 0; i <= 1000; i++)
    for (j = 0; j <= 1000; j++)
      if (a[i][j].size() > 0) {
        curx = i;
        cury = j;
        print(i, j);
        j = 1002;
        i = 1002;
      }
  pair<long long, long long> temp = nearest(curx, cury);
  pair<long long, long long> nothing = make_pair(-1, -1);
  curx = temp.first;
  cury = temp.second;
  while (temp != nothing) {
    print(curx, cury);
    temp = nearest(curx, cury);
    curx = temp.first;
    cury = temp.second;
  }
  return 0;
}
