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:102400000,102400000")
using namespace std;
const int prime = 999983;
const int INF = 0x3f3f3f3f;
const int INFF = 1e9;
const double pi = 3.141592653589793;
const double inf = 1e18;
const double eps = 1e-6;
const int mod = 1e9 + 7;
inline int read_int() {
  int ret = 0;
  char tmp;
  while (!isdigit(tmp = getchar()))
    ;
  do {
    ret = (ret << 3) + (ret << 1) + tmp - '0';
  } while (isdigit(tmp = getchar()));
  return ret;
}
struct Node {
  int x, y;
  int id, d;
  bool operator<(const Node &a) const {
    if (id == a.id && id % 2 == 1) return y > a.y;
    if (id == a.id && id % 2 == 0) return y < a.y;
    return id < a.id;
  }
} node[1000005];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      node[i] = (Node){a, b, a / 1000, i + 1};
    }
    sort(node, node + n);
    for (int i = 0; i < n; i++) {
      printf("%d", node[i].d);
      if (i == n - 1)
        printf("\n");
      else
        printf(" ");
    }
  }
  return 0;
}
