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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
vector<pair<pair<int, int>, int> > pnt;
map<pair<int, int>, int> idx;
void swap(pair<int, int> &a, pair<int, int> &b) {
  pair<int, int> tmp = a;
  a = b;
  b = tmp;
}
bool cmp(const pair<pair<int, int>, int> &a,
         const pair<pair<int, int>, int> &b) {
  return a.first.first < b.first.first ||
         (a.first.first == b.first.first &&
          (a.first.first % 2 == 0 ? (a.first.second < b.first.second)
                                  : (a.first.second > b.first.second)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int N;
  cin >> N;
  for (int i = 0; i < N; ++i) {
    int x, y;
    cin >> x >> y;
    x /= 1000;
    pnt.push_back(pair<pair<int, int>, int>(pair<int, int>(x, y), i));
  }
  sort(pnt.begin(), pnt.end(), cmp);
  for (int i = 0; i < N; ++i) {
    if (i) cout << " ";
    cout << (pnt[i].second + 1);
  }
  cout << endl;
  return 0;
}
