#include <bits/stdc++.h>
using namespace std;
const unsigned long long MAXN = 2 * 1e5 + 10, MAXM = 13;
const unsigned long long L_INF = 1e18 * 4;
const int INF = 1e9;
vector<int> mas[810][810];
bool used[810][810];
void f(int x, int y) {
  used[x][y] = true;
  for (auto a : mas[x][y]) {
    printf("%i ", a + 1);
  }
  if (x < 800 && !used[x + 1][y]) f(x + 1, y);
  if (y < 800 && !used[x][y + 1]) f(x, y + 1);
  if (y < 800 && x < 800 && !used[x + 1][y + 1]) f(x + 1, y + 1);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0, x, y; i < n; i++) {
    scanf("%i%i", &x, &y);
    mas[x / 1250][y / 1250].push_back(i);
  }
  f(0, 0);
}

### Prompt

Please create a solution in Cpp 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 unsigned long long MAXN = 2 * 1e5 + 10, MAXM = 13;
const unsigned long long L_INF = 1e18 * 4;
const int INF = 1e9;
vector<int> mas[810][810];
bool used[810][810];
void f(int x, int y) {
  used[x][y] = true;
  for (auto a : mas[x][y]) {
    printf("%i ", a + 1);
  }
  if (x < 800 && !used[x + 1][y]) f(x + 1, y);
  if (y < 800 && !used[x][y + 1]) f(x, y + 1);
  if (y < 800 && x < 800 && !used[x + 1][y + 1]) f(x + 1, y + 1);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0, x, y; i < n; i++) {
    scanf("%i%i", &x, &y);
    mas[x / 1250][y / 1250].push_back(i);
  }
  f(0, 0);
}

```