#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
const long long INF = (1LL << 60);
const int N = 1001000;
struct Point {
  int id, blockid, l, r;
  bool operator<(const Point &b) const {
    if (blockid != b.blockid) {
      return blockid < b.blockid;
    }
    if (blockid & 1) {
      return b.r < r;
    }
    return r < b.r;
  }
} p[N];
int main() {
  int n;
  int bsize = 1000;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &p[i].l, &p[i].r);
    p[i].blockid = p[i].l / bsize;
    p[i].id = i;
  }
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; ++i) {
    printf("%d ", p[i].id);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
const long long INF = (1LL << 60);
const int N = 1001000;
struct Point {
  int id, blockid, l, r;
  bool operator<(const Point &b) const {
    if (blockid != b.blockid) {
      return blockid < b.blockid;
    }
    if (blockid & 1) {
      return b.r < r;
    }
    return r < b.r;
  }
} p[N];
int main() {
  int n;
  int bsize = 1000;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &p[i].l, &p[i].r);
    p[i].blockid = p[i].l / bsize;
    p[i].id = i;
  }
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; ++i) {
    printf("%d ", p[i].id);
  }
  return 0;
}

```