#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x;
  int y;
  int id;
  Point() {}
  Point(int a, int b, int d) {
    x = a;
    y = b;
    id = d;
  }
};
bool ordY(Point a, Point b) {
  if (a.y == b.y) return a.x < b.x;
  return a.y < b.y;
}
bool ordX(Point a, Point b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
Point read(int i) {
  int x, y;
  cin >> x >> y;
  return Point(x, y, i);
}
int main() {
  int n;
  cin >> n;
  Point p;
  vector<Point> ps[1001];
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &p.x, &p.y);
    p.id = i + 1;
    ps[(p.x + 1) / 1000].push_back(p);
  }
  for (int i = 0; i <= 1000; i++) {
    sort(ps[i].begin(), ps[i].end(), ordY);
    for (int j = 0; j < ps[i].size(); j++) {
      cout << ps[i][j].id << " ";
    }
  }
  cout << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
struct Point {
  int x;
  int y;
  int id;
  Point() {}
  Point(int a, int b, int d) {
    x = a;
    y = b;
    id = d;
  }
};
bool ordY(Point a, Point b) {
  if (a.y == b.y) return a.x < b.x;
  return a.y < b.y;
}
bool ordX(Point a, Point b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
Point read(int i) {
  int x, y;
  cin >> x >> y;
  return Point(x, y, i);
}
int main() {
  int n;
  cin >> n;
  Point p;
  vector<Point> ps[1001];
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &p.x, &p.y);
    p.id = i + 1;
    ps[(p.x + 1) / 1000].push_back(p);
  }
  for (int i = 0; i <= 1000; i++) {
    sort(ps[i].begin(), ps[i].end(), ordY);
    for (int j = 0; j < ps[i].size(); j++) {
      cout << ps[i][j].id << " ";
    }
  }
  cout << endl;
  return 0;
}

```