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;
struct pairhash {
 public:
  template <typename T, typename U>
  size_t operator()(const pair<T, U> &x) const {
    size_t seed = hash<T>()(x.first);
    return hash<U>()(x.second) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
  }
};
const int inf = 1e9 + 9;
const long long mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1);
unsigned long xor128(void) {
  static unsigned long x = 123456789, y = 362436069, z = 521288629,
                       w = 88675123;
  unsigned long t;
  t = (x ^ (x << 11));
  x = y;
  y = z;
  z = w;
  return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
int rand(int k) { return xor128() % k; }
int n;
int x[1000010], y[1000010];
inline int dist(int a, int b) { return abs(x[a] - x[b]) + abs(y[a] - y[b]); }
void solve() {
  vector<int> res;
  vector<int> p[1001];
  for (int i = 0; i < n; i++) {
    p[x[i] / 1000].push_back(i);
  }
  for (int i = 0; i <= 1000; i++) {
    if (i % 2 == 0) {
      sort((p[i]).begin(), (p[i]).end(), [&](const int &a, const int &b) {
        return y[a] < y[b] || (y[a] == y[b] && x[a] < x[b]);
      });
    } else {
      sort((p[i]).begin(), (p[i]).end(), [&](const int &a, const int &b) {
        return y[a] > y[b] || (y[a] == y[b] && x[a] < x[b]);
      });
    }
    for (int j = 0; j < p[i].size(); j++) res.push_back(p[i][j]);
  }
  for (int i = 0; i < n; i++) {
    printf("%d", res[i] + 1);
    if (i < n - 1)
      printf(" ");
    else
      printf("\n");
  }
}
void input() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
  }
}
int main() {
  input();
  solve();
}
