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 powe(long long n, long long m) {
  if (m == 0) return 1;
  long long t = powe(n, m / 2);
  if (m % 2 == 0) return (t * t);
  return (((t * t)) * n);
}
long long mpowe(long long n, long long m, long long MOD) {
  if (m == 0) return 1;
  long long t = mpowe(n, m / 2, MOD);
  t %= MOD;
  if (m % 2 == 0) return (t * t) % MOD;
  return (((t * t) % MOD) * n) % MOD;
}
long long logtwo(long long n) {
  if (n == 1) return 0;
  if (n == 0) return -1;
  return logtwo(n / 2) + 1;
}
bool compare(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / 1200 == b.first.first / 1200) {
    if ((a.first.first / 1200) % 2) return a.first.second < b.first.second;
    return a.first.second > b.first.second;
  } else {
    return a.first.first < b.first.first;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  std::vector<pair<pair<int, int>, int>> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first.first >> v[i].first.second;
    v[i].second = i + 1;
  }
  sort(v.begin(), v.end(), compare);
  for (int i = 0; i < n; i++) {
    cout << v[i].second << " ";
  }
}
