#include <bits/stdc++.h>
using namespace std;
inline int Read() {
  int x;
  scanf("%d", &x);
  return x;
}
pair<pair<int, int>, int> P[1000000];
int n;
bool key(pair<pair<int, int>, int> X, pair<pair<int, int>, int> Y) {
  if (X.first.first / 1000 != Y.first.first / 1000)
    return X.first.first < Y.first.first;
  else
    return X.first.second < Y.first.second;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    P[i - 1] = pair<pair<int, int>, int>(pair<int, int>(x, y), i);
  }
  sort(P, P + n, key);
  int tmp = 0, l = -1;
  while (l + 1 < n) {
    ++l;
    int r = l;
    while (r + 1 < n && P[r + 1].first.first / 1000 == P[l].first.first / 1000)
      ++r;
    if (tmp % 2 == 0)
      for (int i = (l); i <= (r); ++i) printf("%d ", P[i].second);
    else
      for (int i = (r); i >= (l); --i) printf("%d ", P[i].second);
    tmp = 1 - tmp;
    l = r;
  }
  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;
inline int Read() {
  int x;
  scanf("%d", &x);
  return x;
}
pair<pair<int, int>, int> P[1000000];
int n;
bool key(pair<pair<int, int>, int> X, pair<pair<int, int>, int> Y) {
  if (X.first.first / 1000 != Y.first.first / 1000)
    return X.first.first < Y.first.first;
  else
    return X.first.second < Y.first.second;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    P[i - 1] = pair<pair<int, int>, int>(pair<int, int>(x, y), i);
  }
  sort(P, P + n, key);
  int tmp = 0, l = -1;
  while (l + 1 < n) {
    ++l;
    int r = l;
    while (r + 1 < n && P[r + 1].first.first / 1000 == P[l].first.first / 1000)
      ++r;
    if (tmp % 2 == 0)
      for (int i = (l); i <= (r); ++i) printf("%d ", P[i].second);
    else
      for (int i = (r); i >= (l); --i) printf("%d ", P[i].second);
    tmp = 1 - tmp;
    l = r;
  }
  return 0;
}

```