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;
inline char gc() {
  static char buf[1 << 16], *S, *T;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline int read() {
  int x = 0, f = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = gc();
  return x * f;
}
int n;
struct point {
  int x, y, id;
} P[1000010];
inline bool cmp(point a, point b) { return a.x < b.x; }
inline bool cmp1(point a, point b) { return a.y < b.y; }
inline bool cmp2(point a, point b) { return a.y > b.y; }
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) P[i].x = read(), P[i].y = read(), P[i].id = i;
  sort(P + 1, P + n + 1, cmp);
  for (int i = 1, now = 1; i <= n; ++i, ++now) {
    int last = i;
    while (last <= n && P[last].x <= now * 1000) ++last;
    if (now & 1)
      sort(P + i, P + last, cmp1);
    else
      sort(P + i, P + last, cmp2);
    for (int j = i; j < last; ++j) printf("%d ", P[j].id);
    i = last - 1;
  }
  return 0;
}
