#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 100;
int n;
vector<int> block[1100][1100];
vector<pair<int, int> > pts;
bool cmpl(int a, int b) { return pts[a].first < pts[b].first; }
bool cmpg(int a, int b) { return pts[a].first > pts[b].first; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  pts.push_back({0, 0});
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    block[a / 1000][b / 1000].push_back(i + 1);
    pts.push_back({a, b});
  }
  vector<int> res;
  for (int i = 0; i <= 1000; i++) {
    if (i % 2 == 0)
      for (int j = 0; j <= 1000; j++) {
        if ((int)block[i][j].size()) {
          sort(block[i][j].begin(), block[i][j].end(), cmpl);
          for (auto aa : block[i][j]) res.push_back(aa);
        }
      }
    else
      for (int j = 1000; j >= 0; j--) {
        if ((int)block[i][j].size()) {
          sort(block[i][j].begin(), block[i][j].end(), cmpg);
          for (auto aa : block[i][j]) res.push_back(aa);
        }
      }
  }
  for (auto aa : res) cout << aa << " ";
  cout << "\n";
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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 = 5e5 + 100;
int n;
vector<int> block[1100][1100];
vector<pair<int, int> > pts;
bool cmpl(int a, int b) { return pts[a].first < pts[b].first; }
bool cmpg(int a, int b) { return pts[a].first > pts[b].first; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  pts.push_back({0, 0});
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    block[a / 1000][b / 1000].push_back(i + 1);
    pts.push_back({a, b});
  }
  vector<int> res;
  for (int i = 0; i <= 1000; i++) {
    if (i % 2 == 0)
      for (int j = 0; j <= 1000; j++) {
        if ((int)block[i][j].size()) {
          sort(block[i][j].begin(), block[i][j].end(), cmpl);
          for (auto aa : block[i][j]) res.push_back(aa);
        }
      }
    else
      for (int j = 1000; j >= 0; j--) {
        if ((int)block[i][j].size()) {
          sort(block[i][j].begin(), block[i][j].end(), cmpg);
          for (auto aa : block[i][j]) res.push_back(aa);
        }
      }
  }
  for (auto aa : res) cout << aa << " ";
  cout << "\n";
  return 0;
}

```