#include <bits/stdc++.h>
using namespace std;
struct node {
  int x, y;
  int index;
} nod[1000010];
vector<node> vec[1050];
bool cmp(node a, node b) { return a.y < b.y; }
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < 1050; i++) {
      vec[i].clear();
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &nod[i].x, &nod[i].y);
      nod[i].index = i;
      vec[nod[i].x / 1000].push_back(nod[i]);
    }
    int flag = 0;
    for (int i = 0; i <= 1000; i++) {
      if (vec[i].size() > 0) {
        sort(vec[i].begin(), vec[i].end(), cmp);
        if (i & 1) {
          for (int j = 0; j < vec[i].size(); j++) {
            if (flag) {
              printf(" %d", vec[i][j].index);
            } else {
              printf("%d", vec[i][j].index);
              flag = 1;
            }
          }
        } else {
          for (int j = vec[i].size() - 1; j >= 0; j--) {
            if (flag) {
              printf(" %d", vec[i][j].index);
            } else {
              printf("%d", vec[i][j].index);
              flag = 1;
            }
          }
        }
      }
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
struct node {
  int x, y;
  int index;
} nod[1000010];
vector<node> vec[1050];
bool cmp(node a, node b) { return a.y < b.y; }
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < 1050; i++) {
      vec[i].clear();
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &nod[i].x, &nod[i].y);
      nod[i].index = i;
      vec[nod[i].x / 1000].push_back(nod[i]);
    }
    int flag = 0;
    for (int i = 0; i <= 1000; i++) {
      if (vec[i].size() > 0) {
        sort(vec[i].begin(), vec[i].end(), cmp);
        if (i & 1) {
          for (int j = 0; j < vec[i].size(); j++) {
            if (flag) {
              printf(" %d", vec[i][j].index);
            } else {
              printf("%d", vec[i][j].index);
              flag = 1;
            }
          }
        } else {
          for (int j = vec[i].size() - 1; j >= 0; j--) {
            if (flag) {
              printf(" %d", vec[i][j].index);
            } else {
              printf("%d", vec[i][j].index);
              flag = 1;
            }
          }
        }
      }
    }
    cout << endl;
  }
  return 0;
}

```