#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 2, inf = 1e18;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<array<long long, 3>> pts[1002];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x, y;
    cin >> x >> y;
    pts[y / 1000].push_back({x, y, i});
  }
  for (long long i = 0; i <= 1001; i++) {
    if (i & 1)
      sort(pts[i].rbegin(), pts[i].rend());
    else
      sort(pts[i].begin(), pts[i].end());
    for (auto& it : pts[i]) cout << it[2] << ' ';
  }
}

### 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>
using namespace std;
const long long N = 2e5 + 2, inf = 1e18;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<array<long long, 3>> pts[1002];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x, y;
    cin >> x >> y;
    pts[y / 1000].push_back({x, y, i});
  }
  for (long long i = 0; i <= 1001; i++) {
    if (i & 1)
      sort(pts[i].rbegin(), pts[i].rend());
    else
      sort(pts[i].begin(), pts[i].end());
    for (auto& it : pts[i]) cout << it[2] << ' ';
  }
}

```