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 = 1000005;
struct node {
  int x, y, num;
} s[maxn];
bool cmp1(node a, node b) { return a.x < b.x; }
bool cmp2(node a, node b) { return a.y < b.y; }
bool cmp3(node a, node b) { return a.y > b.y; }
int main() {
  int n, i, m;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &s[i].x, &s[i].y);
    s[i].num = i;
  }
  sort(s + 1, s + n + 1, cmp1);
  m = (n % 1000 == 0 ? n / 1000 : n / 1000 + 1);
  for (i = 1; i <= m; i++) {
    if (i & 1)
      sort(s + (i - 1) * 1000 + 1, s + min(i * 1000, n) + 1, cmp2);
    else
      sort(s + (i - 1) * 1000 + 1, s + min(i * 1000, n) + 1, cmp3);
  }
  for (i = 1; i <= n; i++) printf("%d ", s[i].num);
  printf("\n");
  return 0;
}
