#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005;
int S;
int n;
struct ppp {
  int x, y, id, block, mark;
  void read(int _id) {
    scanf("%d%d", &x, &y);
    id = _id;
    block = x / S;
    mark = block % 2;
  }
} node[maxn];
int cmp(ppp a, ppp b) {
  if (a.block != b.block) return a.block < b.block;
  if (a.mark == 0) {
    return a.y > b.y;
  } else
    return a.y < b.y;
}
int main() {
  while (~scanf("%d", &n)) {
    S = sqrt(n);
    if (S <= 0) S = 1;
    for (int i = 1; i <= n; i++) node[i].read(i);
    sort(node + 1, node + 1 + n, cmp);
    for (int i = 1; i <= n; i++) printf("%d ", node[i].id);
    printf("\n");
  }
}

### 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;
const int maxn = 1000005;
int S;
int n;
struct ppp {
  int x, y, id, block, mark;
  void read(int _id) {
    scanf("%d%d", &x, &y);
    id = _id;
    block = x / S;
    mark = block % 2;
  }
} node[maxn];
int cmp(ppp a, ppp b) {
  if (a.block != b.block) return a.block < b.block;
  if (a.mark == 0) {
    return a.y > b.y;
  } else
    return a.y < b.y;
}
int main() {
  while (~scanf("%d", &n)) {
    S = sqrt(n);
    if (S <= 0) S = 1;
    for (int i = 1; i <= n; i++) node[i].read(i);
    sort(node + 1, node + 1 + n, cmp);
    for (int i = 1; i <= n; i++) printf("%d ", node[i].id);
    printf("\n");
  }
}

```