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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const unsigned long long prime = 999983;
const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const unsigned long long mxx = 1333331;
inline void RI(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0';
}
struct po {
  int x, y, num;
};
vector<po> v[1010];
bool cmp(po a, po b) {
  if (a.y < b.y)
    return true;
  else
    return false;
}
int main() {
  int n, x, y;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x, &y);
    po temp;
    temp.num = i;
    temp.x = x;
    temp.y = y;
    v[x / 1000].push_back(temp);
  }
  for (int i = 0; i < 1002; i++) sort(v[i].begin(), v[i].end(), cmp);
  int bool1 = 1;
  for (int i = 0; i < 1002; i++) {
    if (bool1) {
      for (int j = 0; j < v[i].size(); j++) {
        printf("%d ", v[i][j].num);
      }
    } else {
      for (int j = v[i].size() - 1; j >= 0; j--) {
        printf("%d ", v[i][j].num);
      }
    }
    bool1 = (bool1 + 1) % 2;
  }
  return 0;
}
