#include <bits/stdc++.h>
using namespace std;
void splitstr(const string &s, vector<string> &v) {
  istringstream in(s);
  copy(istream_iterator<string>(in), istream_iterator<string>(),
       back_inserter(v));
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << *it << " = " << a << '\n';
  err(++it, args...);
}
const int maxn = 20005;
const long long INF = (1e9) + 7;
const long double EPS = 1e-8;
vector<pair<pair<int, int>, int> > a[maxn];
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (int)n; i++) {
    int x, y;
    cin >> x >> y;
    int p = y / 1300;
    a[p].push_back(make_pair(make_pair(x, y), i + 1));
  }
  for (int i = 0; i < (int)maxn; i++) sort(a[i].begin(), a[i].end());
  for (int i = 0; i < (int)maxn; i++)
    for (int j = 0; j < (int)a[i].size(); j++) ans.push_back(a[i][j].second);
  for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << " ";
  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;
void splitstr(const string &s, vector<string> &v) {
  istringstream in(s);
  copy(istream_iterator<string>(in), istream_iterator<string>(),
       back_inserter(v));
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << *it << " = " << a << '\n';
  err(++it, args...);
}
const int maxn = 20005;
const long long INF = (1e9) + 7;
const long double EPS = 1e-8;
vector<pair<pair<int, int>, int> > a[maxn];
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (int)n; i++) {
    int x, y;
    cin >> x >> y;
    int p = y / 1300;
    a[p].push_back(make_pair(make_pair(x, y), i + 1));
  }
  for (int i = 0; i < (int)maxn; i++) sort(a[i].begin(), a[i].end());
  for (int i = 0; i < (int)maxn; i++)
    for (int j = 0; j < (int)a[i].size(); j++) ans.push_back(a[i][j].second);
  for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << " ";
  return 0;
}

```