#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
int ans = 0, a[1000000], s[1000000], d[1000000], h[1000000];
void l(int e, int e2, int r, int r2, int v) {
  if (v == -1) return;
  if (d[v] == -1) {
    h[ans] = v;
    ans++;
    return;
  }
  int c, e1 = e + ((e2 - e) >> 1), r1 = r + ((r2 - r) >> 1);
  int w, t1 = -1, t2 = -1, t3 = -1, t4 = -1;
  for (w = v; w >= 0;) {
    c = d[w];
    if (a[w] <= e1)
      if (s[w] <= r1) {
        d[w] = t1;
        t1 = w;
      } else {
        d[w] = t4;
        t4 = w;
      }
    else if (s[w] <= r1) {
      d[w] = t2;
      t2 = w;
    } else {
      d[w] = t3;
      t3 = w;
    }
    w = c;
  }
  l(e, e1, r, r1, t1);
  l(e1 + 1, e2, r, r1, t2);
  l(e, e1, r1 + 1, r2, t4);
  l(e1 + 1, e2, r1 + 1, r2, t3);
}
int main() {
  int q, w, e, r, t;
  cin >> q;
  for (w = 0; w < q; w++) {
    scanf("%d%d", &a[w], &s[w]);
    d[w] = w + 1;
  }
  d[q - 1] = -1;
  l(0, 1000000, 0, 1000000, 0);
  for (w = 0; w < q; w++) {
    printf("%d ", h[w] + 1);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#pragma warning(disable : 4996)
int ans = 0, a[1000000], s[1000000], d[1000000], h[1000000];
void l(int e, int e2, int r, int r2, int v) {
  if (v == -1) return;
  if (d[v] == -1) {
    h[ans] = v;
    ans++;
    return;
  }
  int c, e1 = e + ((e2 - e) >> 1), r1 = r + ((r2 - r) >> 1);
  int w, t1 = -1, t2 = -1, t3 = -1, t4 = -1;
  for (w = v; w >= 0;) {
    c = d[w];
    if (a[w] <= e1)
      if (s[w] <= r1) {
        d[w] = t1;
        t1 = w;
      } else {
        d[w] = t4;
        t4 = w;
      }
    else if (s[w] <= r1) {
      d[w] = t2;
      t2 = w;
    } else {
      d[w] = t3;
      t3 = w;
    }
    w = c;
  }
  l(e, e1, r, r1, t1);
  l(e1 + 1, e2, r, r1, t2);
  l(e, e1, r1 + 1, r2, t4);
  l(e1 + 1, e2, r1 + 1, r2, t3);
}
int main() {
  int q, w, e, r, t;
  cin >> q;
  for (w = 0; w < q; w++) {
    scanf("%d%d", &a[w], &s[w]);
    d[w] = w + 1;
  }
  d[q - 1] = -1;
  l(0, 1000000, 0, 1000000, 0);
  for (w = 0; w < q; w++) {
    printf("%d ", h[w] + 1);
  }
  return 0;
}

```