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 IntMaxVal = (int)1e20;
int IntMinVal = (int)-1e20;
long long LongMaxVal = (long long)1e20;
long long LongMinVal = (long long)-1e20;
template <typename T>
inline void minimize(T& a, T b) {
  a = std::min(a, b);
}
template <typename T>
inline void maximize(T& a, T b) {
  a = std::max(a, b);
}
template <typename T>
inline void swap(pair<T, T>& p) {
  swap(p.first, p.second);
}
template <typename T>
struct argument_type;
template <typename T, typename U>
struct argument_type<T(U)> {};
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& s) {
  is >> s.first >> s.second;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (int i = 0; i < v.size(); i++) {
    if (i) os << ' ';
    os << v[i];
  }
  os << '\n';
  return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const vector<pair<T1, T2>>& v) {
  for (int i = 0; i < v.size(); i++) {
    os << v[i] << '\n';
  }
  return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& s, const pair<T1, T2>& t) {
  s << t.first << ' ' << t.second;
  return s;
}
template <typename T>
vector<T> readVector(int n) {
  vector<T> res(n);
  for (int i = 0; i < n; i++) cin >> res[i];
  return res;
}
vector<int> positions;
vector<int> indices;
bool sorter(int x, int y) { return positions[x] < positions[y]; }
inline void get_position(int x, int y) {
  int column = x / 1000;
  if (x - column * 1000 > 500) column++;
  int add = (column & 1) ? 1000000 - y : y;
  positions.push_back(column * 1000000 + column + add);
}
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  scanf("%d", &n);
  positions.reserve(1000000);
  indices.reserve(1000000);
  for (int i = 0; i < n; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    get_position(x, y);
    indices.push_back(i);
  }
  sort(indices.begin(), indices.end(), sorter);
  for (int i = 0; i < n; ++i) printf("%d ", indices[i] + 1);
}
