#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, N = 1e6 + 10, S = 1000;
pair<int, int> A[N];
int n, id[N];
bool cmp(int a, int b) {
  if (A[a].first / S != A[b].first / S) return A[a].first / S < A[b].first / S;
  return (A[a].first / S & 1) ^ (A[a].second < A[b].second);
}
int main() {
  scanf("%d", &n);
  for (int i = (1), i_end = (n); i <= i_end; ++i) {
    id[i] = i;
    scanf("%d%d", &A[i].first, &A[i].second);
  }
  sort(id + 1, id + 1 + n, cmp);
  for (int i = (1), i_end = (n); i <= i_end; ++i) printf("%d ", id[i]);
  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;
const int INF = 0x3f3f3f3f, N = 1e6 + 10, S = 1000;
pair<int, int> A[N];
int n, id[N];
bool cmp(int a, int b) {
  if (A[a].first / S != A[b].first / S) return A[a].first / S < A[b].first / S;
  return (A[a].first / S & 1) ^ (A[a].second < A[b].second);
}
int main() {
  scanf("%d", &n);
  for (int i = (1), i_end = (n); i <= i_end; ++i) {
    id[i] = i;
    scanf("%d%d", &A[i].first, &A[i].second);
  }
  sort(id + 1, id + 1 + n, cmp);
  for (int i = (1), i_end = (n); i <= i_end; ++i) printf("%d ", id[i]);
  return 0;
}

```