#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e7;
long long n;
vector<pair<pair<long long, long long>, long long>> vec;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x, y;
    cin >> x >> y;
    if ((x / 1000) % 2) y = -y;
    vec.push_back({{x / 1000, y}, i});
  }
  sort(vec.begin(), vec.end());
  for (auto i : vec) cout << i.second << ' ';
}

### Prompt

Please create a solution in cpp 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 = 2e5 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e7;
long long n;
vector<pair<pair<long long, long long>, long long>> vec;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x, y;
    cin >> x >> y;
    if ((x / 1000) % 2) y = -y;
    vec.push_back({{x / 1000, y}, i});
  }
  sort(vec.begin(), vec.end());
  for (auto i : vec) cout << i.second << ' ';
}

```