#include <bits/stdc++.h>
using namespace std;
int P[1 << 20][2], n;
vector<int> bucket[1 << 10];
int f;
bool cmp(const int& a, const int& b) {
  if (f)
    return P[a][1] > P[b][1];
  else
    return P[a][1] < P[b][1];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (2); j++) scanf("%d", P[i] + j);
    bucket[P[i][0] / 1000].push_back(i);
  }
  for (int i = 0; i < (1 << 10); i++) {
    sort(bucket[i].begin(), bucket[i].end(), &cmp);
    f = 1 - f;
  }
  bool c = 0;
  for (int i = 0; i < (1 << 10); i++)
    for (auto p : bucket[i]) printf("%s%d", c ? " " : "", p + 1), c = 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;
int P[1 << 20][2], n;
vector<int> bucket[1 << 10];
int f;
bool cmp(const int& a, const int& b) {
  if (f)
    return P[a][1] > P[b][1];
  else
    return P[a][1] < P[b][1];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (2); j++) scanf("%d", P[i] + j);
    bucket[P[i][0] / 1000].push_back(i);
  }
  for (int i = 0; i < (1 << 10); i++) {
    sort(bucket[i].begin(), bucket[i].end(), &cmp);
    f = 1 - f;
  }
  bool c = 0;
  for (int i = 0; i < (1 << 10); i++)
    for (auto p : bucket[i]) printf("%s%d", c ? " " : "", p + 1), c = 1;
  puts("");
  return 0;
}

```