#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 6;
struct Point {
  int x, y, id;
} p[N];
bool cmp(Point p1, Point p2) { return p1.y < p2.y; }
vector<Point> G[1005];
int main() {
  int n, ct;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < 1005; i++) G[i].clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &p[i].x, &p[i].y);
      p[i].id = i;
      G[p[i].x / 1000].push_back(p[i]);
    }
    ct = 0;
    bool go = true;
    for (int i = 0; i < 1005; i++) {
      sort(G[i].begin(), G[i].end(), cmp);
      int len = G[i].size();
      if (go)
        for (int j = 0; j < len; j++) {
          printf("%d", G[i][j].id);
          ct++;
          printf("%c", (n == ct) ? '\n' : ' ');
        }
      else
        for (int j = len - 1; j >= 0; j--) {
          printf("%d", G[i][j].id);
          ct++;
          printf("%c", (n == ct) ? '\n' : ' ');
        }
      (go == true) ? go = false : go = true;
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1e6 + 6;
struct Point {
  int x, y, id;
} p[N];
bool cmp(Point p1, Point p2) { return p1.y < p2.y; }
vector<Point> G[1005];
int main() {
  int n, ct;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < 1005; i++) G[i].clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &p[i].x, &p[i].y);
      p[i].id = i;
      G[p[i].x / 1000].push_back(p[i]);
    }
    ct = 0;
    bool go = true;
    for (int i = 0; i < 1005; i++) {
      sort(G[i].begin(), G[i].end(), cmp);
      int len = G[i].size();
      if (go)
        for (int j = 0; j < len; j++) {
          printf("%d", G[i][j].id);
          ct++;
          printf("%c", (n == ct) ? '\n' : ' ');
        }
      else
        for (int j = len - 1; j >= 0; j--) {
          printf("%d", G[i][j].id);
          ct++;
          printf("%c", (n == ct) ? '\n' : ' ');
        }
      (go == true) ? go = false : go = true;
    }
  }
  return 0;
}

```