#include <bits/stdc++.h>
using namespace std;
void Rd(int &x) {
  static char c;
  x = 0;
  while ((c = getchar()) < 48)
    ;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while ((c = getchar()) > 47);
}
int n, X[1000004], Y[1000004], Z[1000004], P[1000004];
bool Cmp(int a, int b) {
  return Z[a] < Z[b] ||
         (Z[a] == Z[b] && ((Z[a] & 1) ? Y[a] < Y[b] : Y[a] > Y[b]));
}
int main() {
  Rd(n);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i)
    Rd(X[i]), Rd(Y[i]), P[i] = i, Z[i] = X[i] / 1000;
  sort(P, P + n, Cmp);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i)
    printf("%d%c", P[i] + 1, " \n"[i == n - 1]);
  return 0;
}

### 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;
void Rd(int &x) {
  static char c;
  x = 0;
  while ((c = getchar()) < 48)
    ;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while ((c = getchar()) > 47);
}
int n, X[1000004], Y[1000004], Z[1000004], P[1000004];
bool Cmp(int a, int b) {
  return Z[a] < Z[b] ||
         (Z[a] == Z[b] && ((Z[a] & 1) ? Y[a] < Y[b] : Y[a] > Y[b]));
}
int main() {
  Rd(n);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i)
    Rd(X[i]), Rd(Y[i]), P[i] = i, Z[i] = X[i] / 1000;
  sort(P, P + n, Cmp);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i)
    printf("%d%c", P[i] + 1, " \n"[i == n - 1]);
  return 0;
}

```