#include <bits/stdc++.h>
using namespace std;
struct Point {
  int l, r, index, block;
  Point() {}
  Point(int l, int r, int index) : l(l), r(r), index(index) {}
  void get_block(int d) { block = l / d; }
  bool operator<(const Point& p) const {
    if (block != p.block) return block < p.block;
    return r < p.r;
  }
} p[1000005];
int n;
int main() {
  scanf("%d", &n);
  int d = 1500;
  for (int i = 0; i < (int(n)); i++) {
    scanf("%d%d", &p[i].l, &p[i].r);
    p[i].index = i;
    p[i].get_block(d);
  }
  sort(p, p + n);
  for (int i = 0; i < (int(n)); i++) {
    if (i) printf(" ");
    printf("%d", p[i].index + 1);
  }
  puts("");
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct Point {
  int l, r, index, block;
  Point() {}
  Point(int l, int r, int index) : l(l), r(r), index(index) {}
  void get_block(int d) { block = l / d; }
  bool operator<(const Point& p) const {
    if (block != p.block) return block < p.block;
    return r < p.r;
  }
} p[1000005];
int n;
int main() {
  scanf("%d", &n);
  int d = 1500;
  for (int i = 0; i < (int(n)); i++) {
    scanf("%d%d", &p[i].l, &p[i].r);
    p[i].index = i;
    p[i].get_block(d);
  }
  sort(p, p + n);
  for (int i = 0; i < (int(n)); i++) {
    if (i) printf(" ");
    printf("%d", p[i].index + 1);
  }
  puts("");
  return 0;
}

```