#include <bits/stdc++.h>
using namespace std;
bool sorta(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return (a.first.second) < (b.first.second);
}
bool sortb(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return (a.first.second) > (b.first.second);
}
int main() {
  vector<vector<pair<pair<int, int>, int> > > vec;
  for (int i = 0; i < 1001; ++i) {
    vector<pair<pair<int, int>, int> > t;
    vec.push_back(t);
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    pair<pair<int, int>, int> coords = make_pair(make_pair(x, y), i);
    int nx = x / 1000;
    vec[nx].push_back(coords);
  }
  for (int i = 0; i < 1001; ++i) {
    if (i % 2) {
      sort(vec[i].begin(), vec[i].end(), sorta);
    } else {
      sort(vec[i].begin(), vec[i].end(), sortb);
    }
  }
  for (int i = 0; i < 1001; ++i) {
    for (int k = 0; k < vec[i].size(); ++k) {
      cout << vec[i][k].second << " ";
    }
  }
  cout << endl;
}

### 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;
bool sorta(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return (a.first.second) < (b.first.second);
}
bool sortb(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return (a.first.second) > (b.first.second);
}
int main() {
  vector<vector<pair<pair<int, int>, int> > > vec;
  for (int i = 0; i < 1001; ++i) {
    vector<pair<pair<int, int>, int> > t;
    vec.push_back(t);
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    pair<pair<int, int>, int> coords = make_pair(make_pair(x, y), i);
    int nx = x / 1000;
    vec[nx].push_back(coords);
  }
  for (int i = 0; i < 1001; ++i) {
    if (i % 2) {
      sort(vec[i].begin(), vec[i].end(), sorta);
    } else {
      sort(vec[i].begin(), vec[i].end(), sortb);
    }
  }
  for (int i = 0; i < 1001; ++i) {
    for (int k = 0; k < vec[i].size(); ++k) {
      cout << vec[i][k].second << " ";
    }
  }
  cout << endl;
}

```