#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
struct point {
  long long x, y, i;
};
bool cmp(const point &i, const point &j) {
  long long blockI = i.x / 1000;
  long long blockJ = j.x / 1000;
  if (blockI == blockJ) {
    if (blockI % 2 == 0) return i.y < j.y;
    return i.y > j.y;
  }
  return blockI < blockJ;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<point> arr(n);
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i].x >> arr[i].y;
    arr[i].i = i + 1;
  }
  sort(arr.begin(), arr.end(), cmp);
  for (long long i = 0; i < n; ++i) {
    cout << arr[i].i << " ";
  }
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
struct point {
  long long x, y, i;
};
bool cmp(const point &i, const point &j) {
  long long blockI = i.x / 1000;
  long long blockJ = j.x / 1000;
  if (blockI == blockJ) {
    if (blockI % 2 == 0) return i.y < j.y;
    return i.y > j.y;
  }
  return blockI < blockJ;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<point> arr(n);
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i].x >> arr[i].y;
    arr[i].i = i + 1;
  }
  sort(arr.begin(), arr.end(), cmp);
  for (long long i = 0; i < n; ++i) {
    cout << arr[i].i << " ";
  }
}

```