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 d = 1000;
pair<int, int> a[1000100];
vector<int> block[1000100];
int n;
bool compLeft(const int &x, const int &y) {
  return (a[x].second < a[y].second);
}
bool compRight(const int &x, const int &y) {
  return (a[x].second > a[y].second);
}
void init(void) {
  scanf("%d", &n);
  for (int i = (1), _b = (n); i <= _b; i = i + 1)
    scanf("%d%d", &a[i].first, &a[i].second);
  for (int i = (1), _b = (n); i <= _b; i = i + 1)
    block[a[i].first / d].push_back(i);
}
void process(void) {
  bool mode = true;
  for (int i = 0, _n = (1000100); i < _n; i = i + 1)
    if (!block[i].empty()) {
      sort((block[i]).begin(), (block[i]).end(), mode ? compLeft : compRight);
      mode ^= 1;
    }
  for (int i = 0, _n = (1000100); i < _n; i = i + 1)
    if (!block[i].empty())
      for (__typeof((block[i]).begin()) it = (block[i]).begin();
           it != (block[i]).end(); it++)
        printf("%d ", *it);
  printf("\n");
}
int main(void) {
  init();
  process();
  return 0;
}
