#include <bits/stdc++.h>
const int MAX = 1e3 + 15;
class point : std::tuple<int, int, int> {
 public:
  int &x() { return std::get<0>(*this); }
  int &y() { return std::get<1>(*this); }
  int &nr() { return std::get<2>(*this); }
};
int main() {
  std::vector<std::vector<point> > columns;
  columns.resize(MAX);
  point p;
  int n;
  scanf("%d", &n) ?: 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &p.x(), &p.y()) ?: 0;
    p.nr() = i;
    columns[p.x() / MAX].push_back(p);
  }
  for (size_t i = 0; i < columns.size(); i++) {
    std::sort(columns[i].begin(), columns[i].end(),
              [i](point &a, point &b) -> bool {
                return i % 2 ? a.y() < b.y() : a.y() > b.y();
              });
  }
  for (auto col : columns) {
    for (point &x : col) {
      printf("%d ", x.nr());
    }
  }
}

### Prompt

Your task is to create 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>
const int MAX = 1e3 + 15;
class point : std::tuple<int, int, int> {
 public:
  int &x() { return std::get<0>(*this); }
  int &y() { return std::get<1>(*this); }
  int &nr() { return std::get<2>(*this); }
};
int main() {
  std::vector<std::vector<point> > columns;
  columns.resize(MAX);
  point p;
  int n;
  scanf("%d", &n) ?: 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &p.x(), &p.y()) ?: 0;
    p.nr() = i;
    columns[p.x() / MAX].push_back(p);
  }
  for (size_t i = 0; i < columns.size(); i++) {
    std::sort(columns[i].begin(), columns[i].end(),
              [i](point &a, point &b) -> bool {
                return i % 2 ? a.y() < b.y() : a.y() > b.y();
              });
  }
  for (auto col : columns) {
    for (point &x : col) {
      printf("%d ", x.nr());
    }
  }
}

```