#include <bits/stdc++.h>
using namespace std;
vector<pair<pair<int, int>, int> > sects[1000], all;
int main() {
  int n;
  scanf("%d", &n);
  all.resize(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &all[i].first.first, &all[i].first.second);
    all[i].second = i + 1;
  }
  sort(all.begin(), all.end());
  int k = 0;
  int nextX = 1001;
  for (int i = 0; i < n; ++i) {
    if (nextX <= all[i].first.first) {
      ++k;
      nextX += 1000;
    }
    int tmp = all[i].first.second;
    all[i].first.second = all[i].first.first * (k % 2 ? 1 : -1);
    all[i].first.first = tmp;
    sects[k].push_back(all[i]);
  }
  k = 0;
  for (int i = 0; i < 1000; ++i) {
    sort(sects[i].begin(), sects[i].end());
    for (int j = 0; j < sects[i].size(); ++j) {
      all[k++] = sects[i][j];
    }
  }
  printf("%d", all[0].second);
  for (int i = 1; i < n; ++i) printf(" %d", all[i].second);
  printf("\n");
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
vector<pair<pair<int, int>, int> > sects[1000], all;
int main() {
  int n;
  scanf("%d", &n);
  all.resize(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &all[i].first.first, &all[i].first.second);
    all[i].second = i + 1;
  }
  sort(all.begin(), all.end());
  int k = 0;
  int nextX = 1001;
  for (int i = 0; i < n; ++i) {
    if (nextX <= all[i].first.first) {
      ++k;
      nextX += 1000;
    }
    int tmp = all[i].first.second;
    all[i].first.second = all[i].first.first * (k % 2 ? 1 : -1);
    all[i].first.first = tmp;
    sects[k].push_back(all[i]);
  }
  k = 0;
  for (int i = 0; i < 1000; ++i) {
    sort(sects[i].begin(), sects[i].end());
    for (int j = 0; j < sects[i].size(); ++j) {
      all[k++] = sects[i][j];
    }
  }
  printf("%d", all[0].second);
  for (int i = 1; i < n; ++i) printf(" %d", all[i].second);
  printf("\n");
  return 0;
}

```