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 Maxn = int(1e6) + 256;
const int INF = int(1e9) + 7;
struct po {
  int x, y, id;
} t[Maxn];
int n, block = 1000, ix = 1;
vector<po> v;
bool cmp(po a, po b) { return (a.y < b.y); }
bool cmp1(po a, po b) { return (a.x < b.x); }
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> t[i].x >> t[i].y;
    t[i].id = i;
  }
  sort(t + 1, t + n + 1, cmp);
  for (int i = 0; i <= 1000; ++i) {
    int l = block * i, r = block * (i + 1);
    while (t[ix].y >= l && t[ix].y < r && ix <= n) {
      v.push_back(t[ix]);
      ix++;
    }
    sort(v.begin(), v.end(), cmp1);
    if (i & 1) {
      reverse(v.begin(), v.end());
    }
    for (auto it : v) cout << it.id << ' ';
    v.clear();
  }
  return 0;
}
