#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
struct point {
  int x, y, kk, id;
  bool operator<(const point &a) const {
    return kk == a.kk ? y < a.y : kk < a.kk;
  }
} pp[maxn];
bool cmp1(point a, point b) { return a.x < b.x; }
bool cmp2(point a, point b) { return a.kk == b.kk ? a.y < b.y : a.kk < b.kk; }
int n, ku;
int main() {
  cin >> n;
  ku = sqrt(n + 0.0) + 10;
  for (int i = (1); i <= (n); i++) {
    scanf("%d%d", &pp[i].x, &pp[i].y);
    pp[i].id = i;
  }
  sort(pp + 1, pp + 1 + n, cmp1);
  for (int i = (1); i <= (n); i++) {
    pp[i].kk = i / ku;
  }
  sort(pp + 1, pp + 1 + n, cmp2);
  printf("%d", pp[1].id);
  for (int i = (2); i <= (n); i++) printf(" %d", pp[i].id);
  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 maxn = 1000010;
struct point {
  int x, y, kk, id;
  bool operator<(const point &a) const {
    return kk == a.kk ? y < a.y : kk < a.kk;
  }
} pp[maxn];
bool cmp1(point a, point b) { return a.x < b.x; }
bool cmp2(point a, point b) { return a.kk == b.kk ? a.y < b.y : a.kk < b.kk; }
int n, ku;
int main() {
  cin >> n;
  ku = sqrt(n + 0.0) + 10;
  for (int i = (1); i <= (n); i++) {
    scanf("%d%d", &pp[i].x, &pp[i].y);
    pp[i].id = i;
  }
  sort(pp + 1, pp + 1 + n, cmp1);
  for (int i = (1); i <= (n); i++) {
    pp[i].kk = i / ku;
  }
  sort(pp + 1, pp + 1 + n, cmp2);
  printf("%d", pp[1].id);
  for (int i = (2); i <= (n); i++) printf(" %d", pp[i].id);
  return 0;
}

```