#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void mn(T& x, const T& y) {
  if (y < x) x = y;
}
template <typename T>
inline void mx(T& x, const T& y) {
  if (x < y) x = y;
}
template <typename T>
inline int sz(const T& x) {
  return x.size();
}
struct Point {
  int32_t i;
  int32_t x;
  int32_t y;
};
Point pts[1000000];
template <typename It>
inline void solve(It l, It r, int32_t minX, int32_t minY, int32_t size) {
  if (std::distance(l, r) <= 1 || size <= 8) {
    for (Point* p = l; p != r; ++p) {
      std::cout << p->i + 1 << ' ';
    }
    return;
  }
  int32_t midX = minX + size;
  int32_t midY = minY + size;
  auto a = std::partition(l, r, [midX](const Point& p) { return p.x < midX; });
  auto b = std::partition(l, a, [midY](const Point& p) { return p.y < midY; });
  auto c = std::partition(a, r, [midY](const Point& p) { return p.y < midY; });
  solve(l, b, minX, minY, size >> 1);
  solve(b, a, minX, midY, size >> 1);
  solve(c, r, midX, midY, size >> 1);
  solve(a, c, midX, minY, size >> 1);
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  for (int i = 0; i < n; ++i) {
    pts[i].i = i;
    std::cin >> pts[i].x >> pts[i].y;
  }
  solve(pts, pts + n, 0, 0, 1 << 19);
  std::cout << std::endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
template <typename T>
inline void mn(T& x, const T& y) {
  if (y < x) x = y;
}
template <typename T>
inline void mx(T& x, const T& y) {
  if (x < y) x = y;
}
template <typename T>
inline int sz(const T& x) {
  return x.size();
}
struct Point {
  int32_t i;
  int32_t x;
  int32_t y;
};
Point pts[1000000];
template <typename It>
inline void solve(It l, It r, int32_t minX, int32_t minY, int32_t size) {
  if (std::distance(l, r) <= 1 || size <= 8) {
    for (Point* p = l; p != r; ++p) {
      std::cout << p->i + 1 << ' ';
    }
    return;
  }
  int32_t midX = minX + size;
  int32_t midY = minY + size;
  auto a = std::partition(l, r, [midX](const Point& p) { return p.x < midX; });
  auto b = std::partition(l, a, [midY](const Point& p) { return p.y < midY; });
  auto c = std::partition(a, r, [midY](const Point& p) { return p.y < midY; });
  solve(l, b, minX, minY, size >> 1);
  solve(b, a, minX, midY, size >> 1);
  solve(c, r, midX, midY, size >> 1);
  solve(a, c, midX, minY, size >> 1);
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  for (int i = 0; i < n; ++i) {
    pts[i].i = i;
    std::cin >> pts[i].x >> pts[i].y;
  }
  solve(pts, pts + n, 0, 0, 1 << 19);
  std::cout << std::endl;
  return 0;
}

```