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 + 10;
const int maxm = 500 * 500 + 10;
const int inf = 1e9;
const long long mod = 1e9 + 7;
int getint() {
  char c;
  while ((c = getchar()) && !(c >= '0' && c <= '9') && c != '-')
    ;
  int ret = c - '0', sgn = 0;
  if (c == '-') sgn = 1, ret = 0;
  while ((c = getchar()) && c >= '0' && c <= '9') ret = ret * 10 + c - '0';
  if (sgn) ret = -ret;
  return ret;
}
struct node {
  int x, y, id;
  bool operator<(const node a) const {
    if (x != a.x)
      return x < a.x;
    else
      return y < a.y;
  }
} p[maxn];
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    int i;
    for (i = 1; i <= n; i++)
      p[i].x = getint(), p[i].x /= 1050, p[i].y = getint(), p[i].id = i;
    sort(p + 1, p + n + 1);
    for (i = 1; i <= n; i++) cout << p[i].id << " ";
    cout << endl;
  }
  return 0;
}
