#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:268435456")
bool ascending(int i, int j) { return (i < j); }
bool descending(int i, int j) { return (i > j); }
using namespace std;
pair<int, int> P[1000 * 1000 + 100];
vector<int> B[1100];
bool cmp(int i, int j) { return P[i].second < P[j].second; }
bool dcmp(int i, int j) { return P[i].second > P[j].second; }
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int(i) = 0; (i) < (n); (i)++) cin >> P[i].first >> P[i].second;
  for (int(i) = 0; (i) < (n); (i)++) {
    B[P[i].first / 1000].push_back(i);
  }
  for (int(i) = 0; (i) < (1020); (i)++) {
    sort(B[i].begin(), B[i].end(), (i % 2 == 1 ? dcmp : cmp));
  }
  for (int(i) = 0; (i) < (1020); (i)++)
    for (int(j) = 0; (j) < (B[i].size()); (j)++) cout << B[i][j] + 1 << " ";
  cout << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
#pragma comment(linker, "/STACK:268435456")
bool ascending(int i, int j) { return (i < j); }
bool descending(int i, int j) { return (i > j); }
using namespace std;
pair<int, int> P[1000 * 1000 + 100];
vector<int> B[1100];
bool cmp(int i, int j) { return P[i].second < P[j].second; }
bool dcmp(int i, int j) { return P[i].second > P[j].second; }
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int(i) = 0; (i) < (n); (i)++) cin >> P[i].first >> P[i].second;
  for (int(i) = 0; (i) < (n); (i)++) {
    B[P[i].first / 1000].push_back(i);
  }
  for (int(i) = 0; (i) < (1020); (i)++) {
    sort(B[i].begin(), B[i].end(), (i % 2 == 1 ? dcmp : cmp));
  }
  for (int(i) = 0; (i) < (1020); (i)++)
    for (int(j) = 0; (j) < (B[i].size()); (j)++) cout << B[i][j] + 1 << " ";
  cout << endl;
}

```