#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n;
vector<pair<pair<int, int>, int> > v[2010];
pair<pair<int, int>, int> OAO[N];
int main() {
  int a, b;
  scanf("%d", &n);
  for (int t = 0; t < n; t++) {
    scanf("%d%d", &a, &b);
    OAO[t] = make_pair(make_pair(a, b), t + 1);
  }
  sort(OAO, OAO + n);
  for (int t = 0; t < n; t++) {
    int a = (OAO[t].first).first;
    int b = (OAO[t].first).second;
    int index = OAO[t].second;
    int which = b / 980;
    v[which].push_back(make_pair(make_pair(a, b), index));
  }
  int cnt = 0;
  for (int i = 0; i <= 2000; i++) {
    sort(v[i].begin(), v[i].end());
    int size = v[i].size();
    for (int j = 0; j < size; j++) {
      cnt++;
      printf("%d%c", v[i][j].second, cnt == n ? '\n' : ' ');
    }
  }
}

### Prompt

Generate 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 N = 1e6 + 10;
int n;
vector<pair<pair<int, int>, int> > v[2010];
pair<pair<int, int>, int> OAO[N];
int main() {
  int a, b;
  scanf("%d", &n);
  for (int t = 0; t < n; t++) {
    scanf("%d%d", &a, &b);
    OAO[t] = make_pair(make_pair(a, b), t + 1);
  }
  sort(OAO, OAO + n);
  for (int t = 0; t < n; t++) {
    int a = (OAO[t].first).first;
    int b = (OAO[t].first).second;
    int index = OAO[t].second;
    int which = b / 980;
    v[which].push_back(make_pair(make_pair(a, b), index));
  }
  int cnt = 0;
  for (int i = 0; i <= 2000; i++) {
    sort(v[i].begin(), v[i].end());
    int size = v[i].size();
    for (int j = 0; j < size; j++) {
      cnt++;
      printf("%d%c", v[i][j].second, cnt == n ? '\n' : ' ');
    }
  }
}

```