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 = 1e6 + 5;
const long long inf = 0x3f3f3f3f3f3f3f3f;
int N, M, T;
int minx, maxx;
struct node {
  int ge, y, id;
  node() {}
  node(int ge, int y, int id) : ge(ge), y(y), id(id) {}
  bool operator<(const node& t) const {
    return ge < t.ge || ge == t.ge && y < t.y;
  }
} lk[maxn];
vector<pair<int, int> > Vx[maxn];
int h[maxn];
int kk(int x, int y) {
  int m = lower_bound(h, h + 1000, y - 999) - h;
  if (m % 2 == 0) return m * 1000001 + x;
  return m * 1000001 + 1000001 - x;
}
int main() {
  cin >> N;
  minx = inf;
  for (int i = 0, j = 0; i <= 1000000; i += 1000, j++) h[j] = i;
  for (int i = 1; i <= N; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    lk[i] = node(kk(x, y), y, i);
  }
  sort(lk + 1, lk + N + 1);
  for (int i = 1; i <= N; i++) printf("%d ", lk[i].id);
  return 0;
}
