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;
const int M = 1000;
const int MaxN = 1000005;
const int MaxC = 1000005;
int N;
int px[MaxN], py[MaxN];
vector<int> bucks[MaxC / M + 10];
void input() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d%d", &px[i], &py[i]);
  }
}
int main() {
  input();
  for (int i = 0; i < N; i++) {
    bucks[py[i] / M].push_back(i);
  }
  vector<int> order;
  for (int buck = 0; buck <= MaxC / M; buck++) {
    sort(bucks[buck].begin(), bucks[buck].end(),
         [&](int lhs, int rhs) { return (px[lhs] < px[rhs]); });
    if (buck % 2 == 0) {
      for (int v : bucks[buck]) order.push_back(v);
    } else {
      for (int i = (int)bucks[buck].size() - 1; i >= 0; i--) {
        order.push_back(bucks[buck][i]);
      }
    }
  }
  for (int v : order) printf("%d ", v + 1);
  printf("\n");
}
