#include <bits/stdc++.h>
using namespace std;
const int maxx = 1000024;
const int buck = 1101;
const long long mn = 25 * 100000000ll;
vector<int> buckets[buck];
pair<int, int> point[maxx];
long long dist(pair<int, int> a, pair<int, int> b) {
  return abs(a.first - b.first) + abs(a.second - b.second);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int e = 0; e < n; e++) {
    int x, y;
    cin >> x >> y;
    buckets[x / buck].push_back(e);
    point[e] = make_pair(x, y);
  }
  vector<int> ans;
  for (int e = 0; e < buck; e++) {
    if (buckets[e].size() == 0) continue;
    sort(buckets[e].begin(), buckets[e].end(),
         [&](int a, int b) { return point[a].second < point[b].second; });
    for (auto &it : buckets[e]) ans.push_back(it);
  }
  long long d = 0;
  for (int e = 0; e < ans.size() - 1; e++)
    d += dist(point[ans[e]], point[ans[e + 1]]);
  assert(d <= mn);
  for (int e = 0; e < ans.size(); e++)
    cout << ans[e] + 1 << " \n"[e == ans.size() - 1];
  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 maxx = 1000024;
const int buck = 1101;
const long long mn = 25 * 100000000ll;
vector<int> buckets[buck];
pair<int, int> point[maxx];
long long dist(pair<int, int> a, pair<int, int> b) {
  return abs(a.first - b.first) + abs(a.second - b.second);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int e = 0; e < n; e++) {
    int x, y;
    cin >> x >> y;
    buckets[x / buck].push_back(e);
    point[e] = make_pair(x, y);
  }
  vector<int> ans;
  for (int e = 0; e < buck; e++) {
    if (buckets[e].size() == 0) continue;
    sort(buckets[e].begin(), buckets[e].end(),
         [&](int a, int b) { return point[a].second < point[b].second; });
    for (auto &it : buckets[e]) ans.push_back(it);
  }
  long long d = 0;
  for (int e = 0; e < ans.size() - 1; e++)
    d += dist(point[ans[e]], point[ans[e + 1]]);
  assert(d <= mn);
  for (int e = 0; e < ans.size(); e++)
    cout << ans[e] + 1 << " \n"[e == ans.size() - 1];
  return 0;
}

```