#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
int blo[1000005];
bool fun(const pair<pair<int, int>, int> &x,
         const pair<pair<int, int>, int> &y) {
  if (blo[x.first.first] != blo[y.first.first]) {
    return blo[x.first.first] < blo[y.first.first];
  }
  if (blo[x.first.first] % 2) {
    return x.first.second > y.first.second;
  } else {
    return x.first.second < y.first.second;
  }
}
void solve() {
  int n;
  cin >> n;
  int x[n + 3], y[n + 3];
  vector<pair<pair<int, int>, int>> v;
  for (int i = 1; i <= n; i++) {
    cin >> x[i] >> y[i];
    v.push_back({{x[i], y[i]}, i});
  }
  for (int i = 0; i <= 1000000; i++) {
    blo[i] = i / 1000;
  }
  sort(v.begin(), v.end(), fun);
  for (int i = 0; i < n; i++) cout << v[i].second << " ";
  cout << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e5 + 5;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
int blo[1000005];
bool fun(const pair<pair<int, int>, int> &x,
         const pair<pair<int, int>, int> &y) {
  if (blo[x.first.first] != blo[y.first.first]) {
    return blo[x.first.first] < blo[y.first.first];
  }
  if (blo[x.first.first] % 2) {
    return x.first.second > y.first.second;
  } else {
    return x.first.second < y.first.second;
  }
}
void solve() {
  int n;
  cin >> n;
  int x[n + 3], y[n + 3];
  vector<pair<pair<int, int>, int>> v;
  for (int i = 1; i <= n; i++) {
    cin >> x[i] >> y[i];
    v.push_back({{x[i], y[i]}, i});
  }
  for (int i = 0; i <= 1000000; i++) {
    blo[i] = i / 1000;
  }
  sort(v.begin(), v.end(), fun);
  for (int i = 0; i < n; i++) cout << v[i].second << " ";
  cout << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

```