#include <bits/stdc++.h>
using namespace std;
vector<long long> P[1100][1100];
long long S = 1000, n;
void work(vector<long long> &T, long long code) {
  if (!code)
    sort(T.begin(), T.end());
  else
    sort(T.rbegin(), T.rend());
  for (long long v : T) cout << v << " ";
}
void xuly() {
  for (long long i = 0; i <= 1000; ++i) {
    if (i % 2 == 0) {
      for (long long j = 0; j <= 1000; ++j) {
        work(P[i][j], 0);
      }
    } else {
      for (long long j = 1000; j >= 0; --j) {
        work(P[i][j], 1);
      }
    }
  }
}
void nhap() {
  long long x, y;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> x >> y;
    P[x / S][y / S].push_back(i);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  nhap();
  xuly();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
vector<long long> P[1100][1100];
long long S = 1000, n;
void work(vector<long long> &T, long long code) {
  if (!code)
    sort(T.begin(), T.end());
  else
    sort(T.rbegin(), T.rend());
  for (long long v : T) cout << v << " ";
}
void xuly() {
  for (long long i = 0; i <= 1000; ++i) {
    if (i % 2 == 0) {
      for (long long j = 0; j <= 1000; ++j) {
        work(P[i][j], 0);
      }
    } else {
      for (long long j = 1000; j >= 0; --j) {
        work(P[i][j], 1);
      }
    }
  }
}
void nhap() {
  long long x, y;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> x >> y;
    P[x / S][y / S].push_back(i);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  nhap();
  xuly();
  return 0;
}

```