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>
const int MAX = 1e3 + 11;
using namespace std;
int N;
vector<pair<int, int>> A[MAX];
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    int x = 0, y = 0;
    scanf("%d %d", &x, &y);
    A[x / MAX].push_back({y, i});
  }
  for (int i = 0; i < MAX; ++i) {
    for (int j = i + 1; j < MAX; ++j) {
      if (A[i].size() < A[j].size()) {
        swap(A[i], A[j]);
      }
    }
  }
  int sz = 0;
  for (int i = 0; i < MAX; ++i) {
    if (!A[i].size()) {
      sz = i + 1;
      break;
    }
  }
  for (int j = 0; j < sz; ++j) {
    if (j & 1) {
      sort(A[j].begin(), A[j].end());
    } else {
      sort(A[j].rbegin(), A[j].rend());
    }
  }
  vector<int> res;
  for (int i = 0; i < sz; ++i) {
    for (int j = 0; j < A[i].size(); ++j) {
      res.push_back(A[i][j].second);
    }
  }
  for (int k = 0; k < res.size(); ++k) {
    printf("%d ", res[k] + 1);
  }
  return 0;
}
