#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793;
const int L = 1000;
int n;
int x[1000010], y[1000010];
int f[1000010];
int ans[1000010];
void init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    f[i] = (x[i] / L);
    if (f[i] % 2)
      f[i] = f[i] * L + y[i] / L;
    else
      f[i] = f[i] * L + L - y[i] / L;
    ans[i] = i;
  }
}
bool cmp(int a, int b) { return f[a] < f[b]; }
void work() {
  sort(ans + 1, ans + n + 1, cmp);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
int main() {
  init();
  work();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 double PI = 3.141592653589793;
const int L = 1000;
int n;
int x[1000010], y[1000010];
int f[1000010];
int ans[1000010];
void init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    f[i] = (x[i] / L);
    if (f[i] % 2)
      f[i] = f[i] * L + y[i] / L;
    else
      f[i] = f[i] * L + L - y[i] / L;
    ans[i] = i;
  }
}
bool cmp(int a, int b) { return f[a] < f[b]; }
void work() {
  sort(ans + 1, ans + n + 1, cmp);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
int main() {
  init();
  work();
  return 0;
}

```