#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
template <class T>
inline bool read(T& n) {
  T x = 0, tmp = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
  if (c == EOF) return false;
  if (c == '-') c = getchar(), tmp = -1;
  while (c >= '0' && c <= '9') x *= 10, x += (c - '0'), c = getchar();
  n = x * tmp;
  return true;
}
template <class T>
inline void write(T n) {
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  int len = 0, data[20];
  while (n) {
    data[len++] = n % 10;
    n /= 10;
  }
  if (!len) data[len++] = 0;
  while (len--) putchar(data[len] + 48);
}
const int MAXN = 1000010;
const int C = 1000;
struct Node {
  int x, y, id;
  bool operator<(const Node& b) const {
    if (x == b.x) {
      if (x & 1)
        return y > b.y;
      else
        return y < b.y;
    }
    return x < b.x;
  }
} a[MAXN];
int main() {
  int n;
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i].x), read(a[i].y);
    a[i].x /= C;
    a[i].id = i;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    write(a[i].id), putchar(' ');
  }
  putchar('\n');
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
template <class T>
inline bool read(T& n) {
  T x = 0, tmp = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
  if (c == EOF) return false;
  if (c == '-') c = getchar(), tmp = -1;
  while (c >= '0' && c <= '9') x *= 10, x += (c - '0'), c = getchar();
  n = x * tmp;
  return true;
}
template <class T>
inline void write(T n) {
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  int len = 0, data[20];
  while (n) {
    data[len++] = n % 10;
    n /= 10;
  }
  if (!len) data[len++] = 0;
  while (len--) putchar(data[len] + 48);
}
const int MAXN = 1000010;
const int C = 1000;
struct Node {
  int x, y, id;
  bool operator<(const Node& b) const {
    if (x == b.x) {
      if (x & 1)
        return y > b.y;
      else
        return y < b.y;
    }
    return x < b.x;
  }
} a[MAXN];
int main() {
  int n;
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i].x), read(a[i].y);
    a[i].x /= C;
    a[i].id = i;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    write(a[i].id), putchar(' ');
  }
  putchar('\n');
  return 0;
}

```