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 dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
inline int nxt() {
  int aaa;
  scanf("%d", &aaa);
  return aaa;
}
inline long long int lxt() {
  long long int aaa;
  scanf("%lld", &aaa);
  return aaa;
}
inline double dxt() {
  double aaa;
  scanf("%lf", &aaa);
  return aaa;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
}
int ar[200010];
struct point {
  int x, y;
  int id;
  point(int x, int y, int id) : x(x), y(y), id(id) {}
  bool operator<(const point& other) const {
    if (x == other.x) return y < other.y;
    return x < other.x;
  }
};
vector<point> v[1005];
int dist(point a, point b) { return abs(a.x - b.x) + abs(a.y - b.y); }
int main() {
  int n = nxt();
  for (int i = 0; i < n; i++) {
    int x = nxt();
    int y = nxt();
    v[y / 1000].push_back(point(x, y, i));
  }
  for (int i = 0; i < 1005; i++) sort(v[i].begin(), v[i].end());
  vector<int> ans;
  point last(0, 0, 0);
  for (int i = 0; i < 1005; i++) {
    if (v[i].size()) {
      if (dist(last, v[i][0]) < dist(last, v[i].back())) {
        for (auto it : v[i]) ans.push_back(it.id), last = it;
      } else {
        for (int j = v[i].size() - 1; j >= 0; j--)
          ans.push_back(v[i][j].id), last = v[i][j];
      }
    }
  }
  for (auto it : ans) printf("%d ", it + 1);
  return 0;
}
