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 double PI = 3.1415926535897932384626433;
const long long mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-12;
template <class T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int maxn = 1e3 + 100;
struct pnt {
  int x, y, id;
  pnt(int xx = 0, int yy = 0, int iid = 0) {
    x = xx;
    y = yy;
    id = iid;
  }
};
bool cmp1(pnt a, pnt b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
bool cmp2(pnt a, pnt b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x > b.x;
}
vector<pnt> vp[maxn];
int n;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    pnt p;
    cin >> p.x >> p.y;
    p.id = i + 1;
    vp[p.y / 1000].push_back(p);
  }
  for (int i = 0; i < maxn; i++) {
    if (i % 2)
      sort(vp[i].begin(), vp[i].end(), cmp1);
    else
      sort(vp[i].begin(), vp[i].end(), cmp2);
  }
  int cnt = 1;
  for (int i = 0; i < maxn; i++) {
    for (int j = 0; j < vp[i].size(); j++) {
      printf("%d%c", vp[i][j].id, (cnt++ == n) ? '\n' : ' ');
    }
  }
  return 0;
}
