#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const int M = 505;
const int nmax = 4e3 + 5;
const long long inf = 1e9;
const int mod = 1e9 + 7;
const int B = 31;
const long long BOUND = 25 * 1e8;
struct node {
  int i;
  int first, second;
  int c;
  int p;
} a[N];
bool f(node a, node b) {
  return (a.c < b.c) ||
         (a.c == b.c && (!a.p ? a.second < b.second : a.second > b.second));
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    int first, second;
    scanf("%d%d", &first, &second);
    int c = first / 1000;
    a[i] = {i + 1, first, second, c, c % 2};
  }
  sort(a, a + n, f);
  for (int i = 0; i < n; ++i) printf("%d ", a[i].i);
  printf("\n");
  return 0;
}

### Prompt

Please formulate a cpp solution to 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 N = 1e6 + 5;
const int M = 505;
const int nmax = 4e3 + 5;
const long long inf = 1e9;
const int mod = 1e9 + 7;
const int B = 31;
const long long BOUND = 25 * 1e8;
struct node {
  int i;
  int first, second;
  int c;
  int p;
} a[N];
bool f(node a, node b) {
  return (a.c < b.c) ||
         (a.c == b.c && (!a.p ? a.second < b.second : a.second > b.second));
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    int first, second;
    scanf("%d%d", &first, &second);
    int c = first / 1000;
    a[i] = {i + 1, first, second, c, c % 2};
  }
  sort(a, a + n, f);
  for (int i = 0; i < n; ++i) printf("%d ", a[i].i);
  printf("\n");
  return 0;
}

```