#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 0x3f3f3f3f, LLINF = 0x3f3f3f3f3f3f3f3f;
using T = tuple<int, int, int>;
const int MN = 1e6 + 1, BS = 1000;
int N;
T pts[MN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  for (auto i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    pts[i] = {x, y, i};
  }
  sort(pts, pts + N, [&](const T &_, const T &__) {
    auto [a, b, ___] = _;
    auto [c, d, ____] = __;
    int b1 = a / BS, b2 = c / BS;
    if (b1 == b2) {
      if (b1 % 2)
        return b > d;
      else
        return b < d;
    }
    return b1 < b2;
  });
  for (auto i = 0; i < N; i++) cout << get<2>(pts[i]) + 1 << " \n"[i == N - 1];
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
using ll = long long;
const ll INF = 0x3f3f3f3f, LLINF = 0x3f3f3f3f3f3f3f3f;
using T = tuple<int, int, int>;
const int MN = 1e6 + 1, BS = 1000;
int N;
T pts[MN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  for (auto i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    pts[i] = {x, y, i};
  }
  sort(pts, pts + N, [&](const T &_, const T &__) {
    auto [a, b, ___] = _;
    auto [c, d, ____] = __;
    int b1 = a / BS, b2 = c / BS;
    if (b1 == b2) {
      if (b1 % 2)
        return b > d;
      else
        return b < d;
    }
    return b1 < b2;
  });
  for (auto i = 0; i < N; i++) cout << get<2>(pts[i]) + 1 << " \n"[i == N - 1];
  return 0;
}

```