#include <bits/stdc++.h>
using namespace std;
inline long long xy2d(int n, int x, int y) {
  int rx, ry;
  long long s;
  long long d = 0;
  for (s = n / 2; s > 0; s /= 2) {
    rx = (x & s) > 0;
    ry = (y & s) > 0;
    d += s * s * ((3 * rx) ^ ry);
    if (ry == 0) {
      if (rx == 1) {
        x = n - 1 - x;
        y = n - 1 - y;
      }
      x = x ^ y;
      y = x ^ y;
      x = x ^ y;
    }
  }
  return d;
}
int N;
int p[1000001];
long long h[1000001];
bool custom(int a, int b) { return h[a] < h[b]; }
int x, y;
int main() {
  N = ceil(log2(1000005));
  N = (1 << N);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &x, &y);
    p[i] = i + 1;
    h[i + 1] = xy2d(N, x, y);
  }
  sort(p, p + n, custom);
  for (int i = 0; i < n; ++i) printf("%d ", p[i]);
}

### 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;
inline long long xy2d(int n, int x, int y) {
  int rx, ry;
  long long s;
  long long d = 0;
  for (s = n / 2; s > 0; s /= 2) {
    rx = (x & s) > 0;
    ry = (y & s) > 0;
    d += s * s * ((3 * rx) ^ ry);
    if (ry == 0) {
      if (rx == 1) {
        x = n - 1 - x;
        y = n - 1 - y;
      }
      x = x ^ y;
      y = x ^ y;
      x = x ^ y;
    }
  }
  return d;
}
int N;
int p[1000001];
long long h[1000001];
bool custom(int a, int b) { return h[a] < h[b]; }
int x, y;
int main() {
  N = ceil(log2(1000005));
  N = (1 << N);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &x, &y);
    p[i] = i + 1;
    h[i + 1] = xy2d(N, x, y);
  }
  sort(p, p + n, custom);
  for (int i = 0; i < n; ++i) printf("%d ", p[i]);
}

```