#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
vector<int> sqrtSort(const vector<pair<int, int> > &q, const int S) {
  int n = q.size();
  vector<long long> pack(n);
  for (int i = 0; i < n; i++) {
    pack[i] =
        (long long)(q[i].first / S) << 50 |
        (long long)((q[i].first / S & 1) == 0 ? q[i].second
                                              : (1 << 25) - 1 - q[i].second)
            << 25 |
        i;
  }
  sort((pack).begin(), (pack).end());
  vector<int> res(n);
  for (int i = 0; i < n; i++) res[i] = pack[i] & ((1 << 25) - 1);
  return res;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    vector<pair<int, int> > p(n);
    for (int(i) = 0; (i) < (int)(n); ++(i))
      scanf("%d%d", &p[i].first, &p[i].second);
    vector<int> ord = sqrtSort(p, 1000);
    for (int(i) = 0; (i) < (int)(n); ++(i)) {
      if (i != 0) putchar(' ');
      printf("%d", ord[i] + 1);
    }
    puts("");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
vector<int> sqrtSort(const vector<pair<int, int> > &q, const int S) {
  int n = q.size();
  vector<long long> pack(n);
  for (int i = 0; i < n; i++) {
    pack[i] =
        (long long)(q[i].first / S) << 50 |
        (long long)((q[i].first / S & 1) == 0 ? q[i].second
                                              : (1 << 25) - 1 - q[i].second)
            << 25 |
        i;
  }
  sort((pack).begin(), (pack).end());
  vector<int> res(n);
  for (int i = 0; i < n; i++) res[i] = pack[i] & ((1 << 25) - 1);
  return res;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    vector<pair<int, int> > p(n);
    for (int(i) = 0; (i) < (int)(n); ++(i))
      scanf("%d%d", &p[i].first, &p[i].second);
    vector<int> ord = sqrtSort(p, 1000);
    for (int(i) = 0; (i) < (int)(n); ++(i)) {
      if (i != 0) putchar(' ');
      printf("%d", ord[i] + 1);
    }
    puts("");
  }
  return 0;
}

```