#include <bits/stdc++.h>
static constexpr int INF = std::numeric_limits<int>::max();
static constexpr int MAXN = 1000000;
static constexpr int SQRT = 1000;
static std::pair<int, int> luvers[MAXN];
int main() {
  int points;
  std::scanf("%d", &points);
  int last_unused = 0;
  static std::vector<int> groups[SQRT + 1][SQRT + 1];
  for (int i = 0; i < points; i++) {
    int x, y;
    std::scanf("%d %d", &x, &y);
    luvers[i].first = x;
    luvers[i].second = y;
    groups[y / SQRT][x / SQRT].push_back(i + 1);
  }
  int curr_row = -1;
  bool direction = false;
  for (int i = 0; i <= SQRT; i++) {
    int start, delta;
    if (i % 2 == 0) {
      start = 0;
      delta = 1;
    } else {
      start = SQRT;
      delta = -1;
    }
    for (int j = start; j <= SQRT && j >= 0; j += delta) {
      std::vector<int>& vec = groups[i][j];
      for (int patel : vec) {
        std::printf("%d ", patel);
      }
    }
  }
  std::cout << std::endl;
  return EXIT_SUCCESS;
}

### 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>
static constexpr int INF = std::numeric_limits<int>::max();
static constexpr int MAXN = 1000000;
static constexpr int SQRT = 1000;
static std::pair<int, int> luvers[MAXN];
int main() {
  int points;
  std::scanf("%d", &points);
  int last_unused = 0;
  static std::vector<int> groups[SQRT + 1][SQRT + 1];
  for (int i = 0; i < points; i++) {
    int x, y;
    std::scanf("%d %d", &x, &y);
    luvers[i].first = x;
    luvers[i].second = y;
    groups[y / SQRT][x / SQRT].push_back(i + 1);
  }
  int curr_row = -1;
  bool direction = false;
  for (int i = 0; i <= SQRT; i++) {
    int start, delta;
    if (i % 2 == 0) {
      start = 0;
      delta = 1;
    } else {
      start = SQRT;
      delta = -1;
    }
    for (int j = start; j <= SQRT && j >= 0; j += delta) {
      std::vector<int>& vec = groups[i][j];
      for (int patel : vec) {
        std::printf("%d ", patel);
      }
    }
  }
  std::cout << std::endl;
  return EXIT_SUCCESS;
}

```