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 int MOD = (int)1e9 + 7;
const int N = 1000005;
const double PI = 4 * atan(1);
const double eps = 1e-7;
const long long oo = 1e10;
const int K = 205;
vector<pair<pair<int, int>, int> > v;
int n;
int batch = 1000;
int sign[N];
int cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / batch != b.first.first / batch) {
    return a.first.first < b.first.first;
  }
  return sign[a.first.first / batch] * a.first.second <
         sign[a.first.first / batch] * b.first.second;
}
set<long long> ss;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v.push_back({{x, y}, i});
    ss.insert(x / batch);
  }
  int pos = 0;
  for (auto t : ss) {
    if (pos % 2 == 0) {
      sign[t] = 1;
    } else {
      sign[t] = -1;
    }
    pos++;
  }
  sort((v).begin(), (v).end(), cmp);
  for (int i = 0; i < n; i++) {
    cout << v[i].second + 1 << " ";
  }
  return 0;
}
