#include <bits/stdc++.h>
using namespace std;
pair<pair<int, int>, int> p[1111111];
bool cmp1(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second < b.first.second;
}
bool cmp2(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second > b.first.second;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &(p[i].first.first), &(p[i].first.second));
    p[i].second = i;
  }
  sort(p + 1, p + n + 1);
  int st = 1, dir = 1;
  p[n + 1] = make_pair(make_pair(0x3f3f3f3f, 0), 0x3f3f3f3f);
  for (int i = 1; i <= n + 1; i++)
    if (p[i].first.first - p[st].first.first > 1000) {
      if (dir == 1)
        sort(p + st, p + i, cmp1);
      else
        sort(p + st, p + i, cmp2);
      for (int j = st; j < i; j++) printf("%d ", p[j].second);
      dir ^= 1;
      st = 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>
using namespace std;
pair<pair<int, int>, int> p[1111111];
bool cmp1(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second < b.first.second;
}
bool cmp2(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second > b.first.second;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &(p[i].first.first), &(p[i].first.second));
    p[i].second = i;
  }
  sort(p + 1, p + n + 1);
  int st = 1, dir = 1;
  p[n + 1] = make_pair(make_pair(0x3f3f3f3f, 0), 0x3f3f3f3f);
  for (int i = 1; i <= n + 1; i++)
    if (p[i].first.first - p[st].first.first > 1000) {
      if (dir == 1)
        sort(p + st, p + i, cmp1);
      else
        sort(p + st, p + i, cmp2);
      for (int j = st; j < i; j++) printf("%d ", p[j].second);
      dir ^= 1;
      st = i;
    }
}

```