#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e6 + 10;
const int BLOCK = 1e3;
struct Point {
  int id;
  int x, y;
  Point() : id(), x(), y() {}
  Point(int _id, int _x, int _y) : id(_id), x(_x), y(_y) {}
  bool operator<(const Point &a) const { return y < a.y; }
  void scan(int _i) {
    id = _i;
    scanf("%d%d", &x, &y);
  }
};
Point p[N];
vector<Point> block[N];
Point ans[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) p[i].scan(i);
  sort(p, p + n);
  for (int i = 0; i < n; i++) block[p[i].x / BLOCK].push_back(p[i]);
  int cnt = 0;
  int prevType = 1;
  for (int i = 0; i < N; i++) {
    if (prevType == 0) reverse(block[i].begin(), block[i].end());
    int maxX = 0, minX = 1e9;
    for (auto x : block[i]) {
      maxX = max(maxX, x.x);
      minX = min(minX, x.x);
      ans[cnt++] = x;
    }
    if (block[i].size() > 0) prevType = 1 - prevType;
  }
  long long len = 0;
  for (int i = 0; i < cnt - 1; i++)
    len += abs(ans[i].x - ans[i + 1].x) + abs(ans[i].y - ans[i + 1].y);
  assert(len <= (long long)25e8);
  for (int i = 0; i < cnt; i++) printf("%d ", ans[i].id + 1);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
const int N = (int)1e6 + 10;
const int BLOCK = 1e3;
struct Point {
  int id;
  int x, y;
  Point() : id(), x(), y() {}
  Point(int _id, int _x, int _y) : id(_id), x(_x), y(_y) {}
  bool operator<(const Point &a) const { return y < a.y; }
  void scan(int _i) {
    id = _i;
    scanf("%d%d", &x, &y);
  }
};
Point p[N];
vector<Point> block[N];
Point ans[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) p[i].scan(i);
  sort(p, p + n);
  for (int i = 0; i < n; i++) block[p[i].x / BLOCK].push_back(p[i]);
  int cnt = 0;
  int prevType = 1;
  for (int i = 0; i < N; i++) {
    if (prevType == 0) reverse(block[i].begin(), block[i].end());
    int maxX = 0, minX = 1e9;
    for (auto x : block[i]) {
      maxX = max(maxX, x.x);
      minX = min(minX, x.x);
      ans[cnt++] = x;
    }
    if (block[i].size() > 0) prevType = 1 - prevType;
  }
  long long len = 0;
  for (int i = 0; i < cnt - 1; i++)
    len += abs(ans[i].x - ans[i + 1].x) + abs(ans[i].y - ans[i + 1].y);
  assert(len <= (long long)25e8);
  for (int i = 0; i < cnt; i++) printf("%d ", ans[i].id + 1);
  return 0;
}

```