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 task();
int main() {
  srand(time(0));
  task();
  return 0;
}
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
const int M = 1000;
struct S {
  pair<int, int> p;
  int id;
};
int n;
S a[N];
int step;
bool cmp(const S &p1, const S &p2) { return p1.p < p2.p; }
bool cmp2(const S &p1, const S &p2) {
  if (p1.p.second != p2.p.second)
    return step % 2 ? p1.p.second < p2.p.second : p1.p.second > p2.p.second;
  return p1.p < p2.p;
}
void task() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &a[i].p.first, &a[i].p.second);
    a[i].id = i + 1;
  }
  sort(a, a + n, cmp);
  for (int i = rand() % 20; i < n; i += M, ++step) {
    sort(a + i, min(a + n, a + i + M), cmp2);
  }
  long long int ans = 0;
  for (int i = 0; i + 1 < n; ++i)
    ans += abs(a[i + 1].p.first - a[i].p.first) +
           abs(a[i + 1].p.second - a[i].p.second);
  (void(1));
  ;
  for (int i = 0; i < n; ++i) printf("%d ", a[i].id);
}
