#include <bits/stdc++.h>
const int bs = 1000;
const int ms = 1001000;
struct PT {
  int x, y;
  int idx;
  PT(int x = 0, int y = 0) {
    this->x = x;
    this->y = y;
  }
  void read() { scanf("%i %i", &x, &y); }
};
bool operator<(PT a, PT b) {
  if (a.x / bs != b.x / bs) {
    return a.x < b.x;
  } else {
    return a.x / bs % 2 == 0 ? a.y < b.y : a.y > b.y;
  }
}
PT pts[ms];
int main() {
  int n;
  std::cin >> n;
  for (int i = 0; i < n; i++) {
    pts[i].read();
    pts[i].idx = i + 1;
  }
  std::sort(pts, pts + n);
  for (int i = 0; i < n; i++) {
    printf("%i%c", pts[i].idx, i + 1 == n ? '\n' : ' ');
  }
}

### 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>
const int bs = 1000;
const int ms = 1001000;
struct PT {
  int x, y;
  int idx;
  PT(int x = 0, int y = 0) {
    this->x = x;
    this->y = y;
  }
  void read() { scanf("%i %i", &x, &y); }
};
bool operator<(PT a, PT b) {
  if (a.x / bs != b.x / bs) {
    return a.x < b.x;
  } else {
    return a.x / bs % 2 == 0 ? a.y < b.y : a.y > b.y;
  }
}
PT pts[ms];
int main() {
  int n;
  std::cin >> n;
  for (int i = 0; i < n; i++) {
    pts[i].read();
    pts[i].idx = i + 1;
  }
  std::sort(pts, pts + n);
  for (int i = 0; i < n; i++) {
    printf("%i%c", pts[i].idx, i + 1 == n ? '\n' : ' ');
  }
}

```