#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int maxN = 1e6;
const long long MOD = 1e9 + 7;
const long long oo = 1e18;
const int S = 1e3;
const int limi = 1e9 + 1;
vector<int> A[S + 1][S + 1];
void Print(vector<int> a) {
  for (int i : a) cout << i << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    int u, v;
    cin >> u >> v;
    A[u / S][v / S].push_back(i);
  }
  for (int i = 0; i <= 1000; i++)
    if (i & 1) {
      for (int j = 0; j <= 1000; j++) Print(A[i][j]);
    } else {
      for (int j = 1000; j >= 0; j--) Print(A[i][j]);
    }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 N = 2e5 + 10;
const int maxN = 1e6;
const long long MOD = 1e9 + 7;
const long long oo = 1e18;
const int S = 1e3;
const int limi = 1e9 + 1;
vector<int> A[S + 1][S + 1];
void Print(vector<int> a) {
  for (int i : a) cout << i << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    int u, v;
    cin >> u >> v;
    A[u / S][v / S].push_back(i);
  }
  for (int i = 0; i <= 1000; i++)
    if (i & 1) {
      for (int j = 0; j <= 1000; j++) Print(A[i][j]);
    } else {
      for (int j = 1000; j >= 0; j--) Print(A[i][j]);
    }
}

```