#include <bits/stdc++.h>
using namespace std;
int INF = 1000000000;
const long long MOD = 1e9 + 7;
vector<pair<int, int> > points;
vector<vector<pair<pair<int, int>, int> > > nums;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  scanf("%d", &n);
  points.resize(n);
  for (int i = 0; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    points[i] = make_pair(x, y);
  }
  nums.resize(1001);
  for (int i = 0; i < n; ++i)
    nums[points[i].second / 1000].push_back(make_pair(points[i], i + 1));
  for (int i = 0; i <= 1000; ++i) sort(nums[i].begin(), nums[i].end());
  int cnt = 0;
  for (int i = 0; i <= 1000; ++i) {
    vector<pair<pair<int, int>, int> >& cur = nums[i];
    if (cur.size() != 0) ++cnt;
    if (cnt % 2) {
      for (int j = 0; j < cur.size(); ++j) printf("%d ", cur[j].second);
    } else {
      for (int j = cur.size() - 1; j >= 0; --j) printf("%d ", cur[j].second);
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int INF = 1000000000;
const long long MOD = 1e9 + 7;
vector<pair<int, int> > points;
vector<vector<pair<pair<int, int>, int> > > nums;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  scanf("%d", &n);
  points.resize(n);
  for (int i = 0; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    points[i] = make_pair(x, y);
  }
  nums.resize(1001);
  for (int i = 0; i < n; ++i)
    nums[points[i].second / 1000].push_back(make_pair(points[i], i + 1));
  for (int i = 0; i <= 1000; ++i) sort(nums[i].begin(), nums[i].end());
  int cnt = 0;
  for (int i = 0; i <= 1000; ++i) {
    vector<pair<pair<int, int>, int> >& cur = nums[i];
    if (cur.size() != 0) ++cnt;
    if (cnt % 2) {
      for (int j = 0; j < cur.size(); ++j) printf("%d ", cur[j].second);
    } else {
      for (int j = cur.size() - 1; j >= 0; --j) printf("%d ", cur[j].second);
    }
  }
  return 0;
}

```