#include <bits/stdc++.h>
using namespace std;
struct pp {
  int x, y;
};
pp A[1000010];
int order[1000010];
int N;
int comp(int a, int b) {
  return (A[a].x == A[b].x ? A[a].y < A[b].y : A[a].x < A[b].x);
}
int compy1(int a, int b) {
  return (A[a].y == A[b].y ? A[a].x < A[b].x : A[a].y < A[b].y);
}
int compy2(int a, int b) {
  return (A[a].y == A[b].y ? A[a].x > A[b].x : A[a].y > A[b].y);
}
int main() {
  cin >> N;
  int i;
  for (i = 0; i < N; ++i) {
    scanf("%d%d", &A[i].x, &A[i].y);
    order[i] = i;
  }
  int s, cur;
  cur = 1000;
  s = 0;
  int b = 0;
  sort(order, order + N, comp);
  for (i = 0; i < N; ++i) {
    if (i < N - 1 && A[order[i + 1]].x < cur) continue;
    if (b)
      sort(order + s, order + i + 1, compy1);
    else
      sort(order + s, order + i + 1, compy2);
    cur += 1000;
    b ^= 1;
    s = i + 1;
  }
  for (i = 0; i < N; ++i) printf("%d ", order[i] + 1);
  return 0;
}

### Prompt

Your task is to create 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;
struct pp {
  int x, y;
};
pp A[1000010];
int order[1000010];
int N;
int comp(int a, int b) {
  return (A[a].x == A[b].x ? A[a].y < A[b].y : A[a].x < A[b].x);
}
int compy1(int a, int b) {
  return (A[a].y == A[b].y ? A[a].x < A[b].x : A[a].y < A[b].y);
}
int compy2(int a, int b) {
  return (A[a].y == A[b].y ? A[a].x > A[b].x : A[a].y > A[b].y);
}
int main() {
  cin >> N;
  int i;
  for (i = 0; i < N; ++i) {
    scanf("%d%d", &A[i].x, &A[i].y);
    order[i] = i;
  }
  int s, cur;
  cur = 1000;
  s = 0;
  int b = 0;
  sort(order, order + N, comp);
  for (i = 0; i < N; ++i) {
    if (i < N - 1 && A[order[i + 1]].x < cur) continue;
    if (b)
      sort(order + s, order + i + 1, compy1);
    else
      sort(order + s, order + i + 1, compy2);
    cur += 1000;
    b ^= 1;
    s = i + 1;
  }
  for (i = 0; i < N; ++i) printf("%d ", order[i] + 1);
  return 0;
}

```