#include <bits/stdc++.h>
using namespace std;
struct point {
  long long x, y;
  int id;
} p[1234560];
int n;
vector<int> lr, ud;
long long cal(point a, point b) { return abs(a.x - b.x) + abs(a.y - b.y); }
bool cmp1(point a, point b) {
  if (a.x != b.x)
    return a.x < b.x;
  else
    return a.y < b.y;
}
bool cmp2(point a, point b) {
  if (a.y != b.y)
    return a.y < b.y;
  else
    return a.x < b.x;
}
bool cmp3(point a, point b) {
  if (a.y != b.y)
    return a.y > b.y;
  else
    return a.x < b.x;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%I64d%I64d", &p[i].x, &p[i].y);
      p[i].id = i + 1;
    }
    sort(p, p + n, cmp1);
    long long sum1 = 0;
    long long pu = 0;
    for (long long i = 0; i < 1000; i++) {
      long long pt = pu;
      while (p[pu].x <= 1000ll * (i + 1) && pu < n) pu++;
      if (i & 1)
        sort(p + pt, p + pu, cmp3);
      else
        sort(p + pt, p + pu, cmp2);
      for (long long j = pt; j < pu; j++) {
        printf("%d ", p[j].id);
      }
    }
    cout << endl;
  }
  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;
struct point {
  long long x, y;
  int id;
} p[1234560];
int n;
vector<int> lr, ud;
long long cal(point a, point b) { return abs(a.x - b.x) + abs(a.y - b.y); }
bool cmp1(point a, point b) {
  if (a.x != b.x)
    return a.x < b.x;
  else
    return a.y < b.y;
}
bool cmp2(point a, point b) {
  if (a.y != b.y)
    return a.y < b.y;
  else
    return a.x < b.x;
}
bool cmp3(point a, point b) {
  if (a.y != b.y)
    return a.y > b.y;
  else
    return a.x < b.x;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%I64d%I64d", &p[i].x, &p[i].y);
      p[i].id = i + 1;
    }
    sort(p, p + n, cmp1);
    long long sum1 = 0;
    long long pu = 0;
    for (long long i = 0; i < 1000; i++) {
      long long pt = pu;
      while (p[pu].x <= 1000ll * (i + 1) && pu < n) pu++;
      if (i & 1)
        sort(p + pt, p + pu, cmp3);
      else
        sort(p + pt, p + pu, cmp2);
      for (long long j = pt; j < pu; j++) {
        printf("%d ", p[j].id);
      }
    }
    cout << endl;
  }
  return 0;
}

```