#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
vector<pair<int, int> > vc[1000001];
pair<int, int> pp[1000001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    int x, y;
    cin >> x >> y;
    vc[x].push_back(make_pair(y, i + 1));
  }
  int fl = 0;
  for (int i = 0; i <= 1000000; i += 1000) {
    int cnt = 0;
    for (int j = i; j < i + 1000; j++) {
      if (j > 1000000) break;
      for (int k = (0); k < (vc[j].size()); ++k) {
        pp[cnt++] = vc[j][k];
      }
    }
    sort(pp, pp + cnt);
    if (fl) reverse(pp, pp + cnt);
    fl ^= 1;
    for (int j = (0); j < (cnt); ++j) cout << pp[j].second << " ";
  }
  cout << endl;
}

### 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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
vector<pair<int, int> > vc[1000001];
pair<int, int> pp[1000001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    int x, y;
    cin >> x >> y;
    vc[x].push_back(make_pair(y, i + 1));
  }
  int fl = 0;
  for (int i = 0; i <= 1000000; i += 1000) {
    int cnt = 0;
    for (int j = i; j < i + 1000; j++) {
      if (j > 1000000) break;
      for (int k = (0); k < (vc[j].size()); ++k) {
        pp[cnt++] = vc[j][k];
      }
    }
    sort(pp, pp + cnt);
    if (fl) reverse(pp, pp + cnt);
    fl ^= 1;
    for (int j = (0); j < (cnt); ++j) cout << pp[j].second << " ";
  }
  cout << endl;
}

```