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 int N = 1000006;
int n;
struct node {
  int x, y, id;
} w[N];
vector<int> ans;
vector<node> block[1005][1005];
bool cmpx(node A, node B) { return A.x < B.x; }
int dis(node A, node B) { return abs(A.x - B.x) + abs(A.y - B.y); }
void output() {
  printf("%d", ans[0]);
  for (int i = 1; i < ans.size(); i++) printf(" %d", ans[i]);
  printf("\n");
}
void Solve() {
  while (~scanf("%d", &n)) {
    ans.clear();
    for (int i = 0; i <= 1000; i++)
      for (int j = 0; j <= 1000; j++) block[i][j].clear();
    for (int i = 1; i <= n; i++) {
      node p;
      scanf("%d%d", &p.x, &p.y);
      p.id = i;
      block[p.x / 1000][p.y / 1000].push_back(p);
    }
    for (int i = 0; i <= 1000; i++) {
      if (i % 2 == 0)
        for (int j = 0; j <= 1000; j++) {
          int len = block[i][j].size();
          if (!len) continue;
          sort(block[i][j].begin(), block[i][j].end(), cmpx);
          for (int k = 0; k < len; k++) ans.push_back(block[i][j][k].id);
        }
      else
        for (int j = 1000; j >= 0; j--) {
          int len = block[i][j].size();
          if (!len) continue;
          sort(block[i][j].begin(), block[i][j].end(), cmpx);
          for (int k = len - 1; k >= 0; k--) ans.push_back(block[i][j][k].id);
        }
    }
    output();
  }
}
int main() {
  Solve();
  return 0;
}
