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 comment(linker, "/STACK:256000000")
using namespace std;
const double eps = 1e-15;
const double pi = acos(-1.0);
const int MAXN = (int)1e6 + 7;
const int INF = (long long)1e9 + 7;
const long long LINF = (long long)4e18;
const int MOD = (long long)1e9 + 7;
const int CNT = 1000;
int solve();
int main() {
  solve();
  return 0;
}
struct pt {
  int x, y, num;
};
vector<pt> a;
vector<int> ans;
bool cmp(pt &p1, pt &p2) {
  return p1.x < p2.x || (p1.x == p2.x && ((p1.x & 1) ^ (p1.y < p2.y)));
}
int solve() {
  int n;
  scanf("%d", &n);
  int x, y;
  for (int i = 0; i < (int)n; i++) {
    scanf("%d%d", &x, &y);
    a.push_back({x / CNT, y, i + 1});
  }
  sort(a.begin(), a.end(), cmp);
  for (pt p : a) {
    printf("%d ", p.num);
  }
  return 0;
}
