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;
struct node {
  int x, y, id;
} a[1000005];
const bool cmp(const node &i, const node &j) { return i.x < j.x; }
const bool cmp1(const node &i, const node &j) { return i.y < j.y; }
vector<node> v;
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  while (cin >> n) {
    ans.clear();
    int x, y;
    for (int i = 1; i <= n; i++) {
      cin >> a[i].x >> a[i].y;
      a[i].id = i;
    }
    int sqrn = sqrt(1000005);
    sort(a + 1, a + n + 1, cmp);
    int j = 1;
    for (int i = sqrn; i <= 1000005; i += sqrn) {
      v.clear();
      while (j <= n && a[j].x <= i) {
        v.push_back(a[j]);
        j++;
      }
      sort(v.begin(), v.end(), cmp1);
      for (auto p : v) {
        ans.push_back(p.id);
      }
    }
    for (auto i : ans) {
      printf("%d ", i);
    }
    printf("\n");
  }
  return 0;
}
