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;
int n;
pair<pair<int, int>, int> pi[(1000007)];
class BUCKET {
 public:
  map<int, vector<int> > mp;
  BUCKET(){};
} buc[(1 << 10)];
bool cmp(const pair<pair<int, int>, int> &x,
         const pair<pair<int, int>, int> &y) {
  return x.first.second < y.first.second;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &pi[i].first.first, &pi[i].first.second);
    pi[i].second = i;
  }
  sort(pi, pi + n, cmp);
  for (int i = 0; i < n; i++) {
    buc[pi[i].first.first / (1 << 10)].mp[pi[i].first.second].push_back(
        pi[i].second);
  }
  bool bu = false, bf = false;
  for (int i = 0; i < (1 << 10); i++) {
    if (bu) {
      for (map<int, vector<int> >::iterator it = buc[i].mp.begin();
           it != buc[i].mp.end(); it++) {
        if (bf) {
          for (int j = 0; j < it->second.size(); j++) {
            printf("%d ", it->second[j] + 1);
          }
          bf = false;
        } else {
          for (int j = it->second.size() - 1; j >= 0; j--) {
            printf("%d ", it->second[j] + 1);
          }
          bf = true;
        }
      }
      bu = false;
    } else {
      for (map<int, vector<int> >::iterator it = buc[i].mp.end();
           it != buc[i].mp.begin(); it--) {
        if (it == buc[i].mp.end()) {
          it--;
        }
        if (bf) {
          for (int j = 0; j < it->second.size(); j++) {
            printf("%d ", it->second[j] + 1);
          }
          bf = false;
        } else {
          for (int j = it->second.size() - 1; j >= 0; j--) {
            printf("%d ", it->second[j] + 1);
          }
          bf = true;
        }
      }
      map<int, vector<int> >::iterator it = buc[i].mp.begin();
      if (buc[i].mp.size() > 1) {
        if (bf) {
          for (int j = 0; j < it->second.size(); j++) {
            printf("%d ", it->second[j] + 1);
          }
          bf = false;
        } else {
          for (int j = it->second.size() - 1; j >= 0; j--) {
            printf("%d ", it->second[j] + 1);
          }
          bf = true;
        }
      }
      bu = true;
    }
  }
  printf("\n");
  return 0;
}
