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>
#pragma GCC optimize("O2,unroll-loops")
using namespace std;
pair<int, int> p[1000006];
int main() {
  int n, x, y;
  scanf("%d", &n);
  vector<pair<pair<int, int>, int> > vec;
  for (int i = 1; i <= n; i++)
    scanf("%d %d", &x, &y), vec.push_back({{x, y}, i}), p[i] = {x, y};
  sort(vec.begin(), vec.end());
  vector<int> res;
  for (int i = 0; i < n;) {
    int x = vec[i].first.first;
    vector<pair<pair<int, int>, int> > temp;
    while (i < n && (vec[i].first.first - x) <= 1000)
      temp.push_back(
          {{vec[i].first.second, vec[i].first.first}, vec[i].second}),
          i++;
    sort(temp.begin(), temp.end());
    for (int j = 0; j < temp.size(); j++) res.push_back(temp[j].second);
  }
  long long dis = 0;
  for (int i = 1; i < n; i++)
    dis += abs(p[res[i]].first - p[res[i - 1]].first) +
           abs(p[res[i]].second - p[res[i - 1]].second);
  long long mx = 2500000000ll;
  if (dis > mx) {
    res.clear();
    for (int i = 0; i < n; i++) swap(vec[i].first.first, vec[i].first.second);
    sort(vec.begin(), vec.end());
    for (int i = 0; i < n;) {
      int x = vec[i].first.first;
      vector<pair<pair<int, int>, int> > temp;
      while (i < n && (vec[i].first.first - x) <= 1000)
        temp.push_back(
            {{vec[i].first.second, vec[i].first.first}, vec[i].second}),
            i++;
      sort(temp.begin(), temp.end());
      for (int j = 0; j < temp.size(); j++) res.push_back(temp[j].second);
    }
  }
  for (int i = 0; i < res.size(); i++) printf("%d ", res[i]);
  puts("");
  return 0;
}
