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
#include <bits/stdc++.h>
using namespace std;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
const long long N = 25 * 1e8, L = 1005;
vector<int> V[L], X, Y;
int cmp(const int a, const int b) { return Y[a] < Y[b]; }
int main() {
  int n;
  RI(n);
  for (int i = 0; i < n; i++) {
    int x, y;
    RI(x, y);
    X.push_back(x);
    Y.push_back(y);
    V[x / 1000].push_back(i);
  }
  for (int i = 0; i < L; i++) {
    sort(V[i].begin(), V[i].end(), cmp);
    if (i & 1) reverse(V[i].begin(), V[i].end());
    for (int id : V[i]) printf("%d ", id + 1);
  }
  return 0;
}
