#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
const int M = 1000;
pair<int, int> a[N];
vector<int> A[N];
int n;
bool cm(int i, int j) { return a[i].second < a[j].second; }
bool cmi(int i, int j) { return a[i].second > a[j].second; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &a[i].first, &a[i].second);
  for (int i = 1; i <= n; ++i) A[a[i].first / M].push_back(i);
  for (int i = 0; i <= M; ++i)
    if (!A[i].empty()) sort(A[i].begin(), A[i].end(), (i % 2) ? cm : cmi);
  for (int i = 0; i <= M; ++i)
    if (!A[i].empty())
      for (vector<int>::iterator j = A[i].begin(); j != A[i].end(); ++j)
        printf("%d ", *j);
  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;
const int N = 1000010;
const int M = 1000;
pair<int, int> a[N];
vector<int> A[N];
int n;
bool cm(int i, int j) { return a[i].second < a[j].second; }
bool cmi(int i, int j) { return a[i].second > a[j].second; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &a[i].first, &a[i].second);
  for (int i = 1; i <= n; ++i) A[a[i].first / M].push_back(i);
  for (int i = 0; i <= M; ++i)
    if (!A[i].empty()) sort(A[i].begin(), A[i].end(), (i % 2) ? cm : cmi);
  for (int i = 0; i <= M; ++i)
    if (!A[i].empty())
      for (vector<int>::iterator j = A[i].begin(); j != A[i].end(); ++j)
        printf("%d ", *j);
  return 0;
}

```