#include <bits/stdc++.h>
using namespace std;
struct Point {
  int x, y, w, i;
  inline bool operator<(const Point &o) const {
    if (w != o.w) return w < o.w;
    return y < o.y;
  }
} v[1000002];
char ch, last;
int n, ns, ans[1000002];
inline void read(int &x) {
  x = ch = 0;
  do {
    last = ch;
    ch = getchar();
  } while (ch < '0' || '9' < ch);
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  if (last == '-') x = -x;
}
int main(void) {
  read(n);
  ns = 1050;
  for (int i = 1; i <= n; i++)
    read(v[i].x), read(v[i].y), v[i].w = v[i].x / ns, v[i].i = i;
  sort(v + 1, v + n + 1);
  for (int i = 1; i < n; i++) printf("%d ", v[i].i);
  printf("%d", v[n].i);
  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;
struct Point {
  int x, y, w, i;
  inline bool operator<(const Point &o) const {
    if (w != o.w) return w < o.w;
    return y < o.y;
  }
} v[1000002];
char ch, last;
int n, ns, ans[1000002];
inline void read(int &x) {
  x = ch = 0;
  do {
    last = ch;
    ch = getchar();
  } while (ch < '0' || '9' < ch);
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  if (last == '-') x = -x;
}
int main(void) {
  read(n);
  ns = 1050;
  for (int i = 1; i <= n; i++)
    read(v[i].x), read(v[i].y), v[i].w = v[i].x / ns, v[i].i = i;
  sort(v + 1, v + n + 1);
  for (int i = 1; i < n; i++) printf("%d ", v[i].i);
  printf("%d", v[n].i);
  return 0;
}

```