#include <bits/stdc++.h>
using namespace std;
const int nmax = 1000;
struct pct {
  int x, y, ind;
  pct() {}
  pct(int _x, int _y, int _ind) : x(_x), y(_y), ind(_ind) {}
  inline bool operator<(const pct &a) const { return (x < a.x); }
};
vector<pct> v[nmax + 1];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[y / nmax].push_back(pct(x, y, i));
  }
  for (int f = 0; f <= nmax; ++f) {
    sort(v[f].begin(), v[f].end());
    if (f % 2 == 1) {
      for (int i = 0; i < (int)v[f].size(); ++i) {
        printf("%d ", v[f][i].ind);
      }
    } else {
      for (int i = (int)v[f].size() - 1; i >= 0; --i) {
        printf("%d ", v[f][i].ind);
      }
    }
  }
  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;
const int nmax = 1000;
struct pct {
  int x, y, ind;
  pct() {}
  pct(int _x, int _y, int _ind) : x(_x), y(_y), ind(_ind) {}
  inline bool operator<(const pct &a) const { return (x < a.x); }
};
vector<pct> v[nmax + 1];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[y / nmax].push_back(pct(x, y, i));
  }
  for (int f = 0; f <= nmax; ++f) {
    sort(v[f].begin(), v[f].end());
    if (f % 2 == 1) {
      for (int i = 0; i < (int)v[f].size(); ++i) {
        printf("%d ", v[f][i].ind);
      }
    } else {
      for (int i = (int)v[f].size() - 1; i >= 0; --i) {
        printf("%d ", v[f][i].ind);
      }
    }
  }
  return 0;
}

```