Problem: For a vector \vec{v} = (x, y), define |v| = √{x^2 + y^2}.

Allen had a bit too much to drink at the bar, which is at the origin. There are n vectors \vec{v_1}, \vec{v_2}, ⋅⋅⋅, \vec{v_n}. Allen will make n moves. As Allen's sense of direction is impaired, during the i-th move he will either move in the direction \vec{v_i} or -\vec{v_i}. In other words, if his position is currently p = (x, y), he will either move to p + \vec{v_i} or p - \vec{v_i}.

Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position p satisfies |p| ≤ 1.5 ⋅ 10^6 so that he can stay safe.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of moves.

Each of the following lines contains two space-separated integers x_i and y_i, meaning that \vec{v_i} = (x_i, y_i). We have that |v_i| ≤ 10^6 for all i.

Output

Output a single line containing n integers c_1, c_2, ⋅⋅⋅, c_n, each of which is either 1 or -1. Your solution is correct if the value of p = ∑_{i = 1}^n c_i \vec{v_i}, satisfies |p| ≤ 1.5 ⋅ 10^6.

It can be shown that a solution always exists under the given constraints.

Examples

Input

3
999999 0
0 999999
999999 0


Output

1 1 -1 


Input

1
-824590 246031


Output

1 


Input

8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899


Output

1 1 1 1 1 1 1 -1 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int max(const int &x, const int &y) { return x > y ? x : y; }
inline int min(const int &x, const int &y) { return x < y ? x : y; }
inline void swap(int &x, int &y) { x ^= y ^= x ^= y; }
inline void chmax(int &x, const int &y) {
  if (x < y) x = y;
}
inline void chmin(int &x, const int &y) {
  if (x > y) x = y;
}
char inputbuf[1 << 23], *p1 = inputbuf, *p2 = inputbuf;
inline int read() {
  int res = 0;
  char ch =
      (p1 == p2 && (p2 = (p1 = inputbuf) + fread(inputbuf, 1, 1 << 21, stdin),
                    p1 == p2)
           ? EOF
           : *p1++);
  bool f = true;
  for (; ch < '0' || ch > '9';
       ch = (p1 == p2 &&
                     (p2 = (p1 = inputbuf) + fread(inputbuf, 1, 1 << 21, stdin),
                     p1 == p2)
                 ? EOF
                 : *p1++))
    if (ch == '-') f = false;
  for (; ch >= '0' && ch <= '9';
       ch = (p1 == p2 &&
                     (p2 = (p1 = inputbuf) + fread(inputbuf, 1, 1 << 21, stdin),
                     p1 == p2)
                 ? EOF
                 : *p1++))
    res = res * 10 + (ch ^ 48);
  return f ? res : -res;
}
const int N = 1e5 + 15;
long long sx, sy;
int n, ans[N];
struct vec {
  long long x, y;
  int id;
  vec(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
} a[N];
inline double dis(const vec &a) { return sqrt(a.x * a.x + a.y * a.y); }
inline double dis(const double &x, const double &y) {
  return sqrt(x * x + y * y);
}
signed main() {
  n = read();
  for (register int i = 1, iend = n; i <= iend; ++i)
    a[i].x = read(), a[i].y = read(), a[i].id = i;
  sx = sy = 1e9;
  while (dis(sx, sy) > 1.5e6) {
    random_shuffle(a + 1, a + n + 1);
    sx = sy = 0;
    for (register int i = 1, iend = n; i <= iend; ++i) {
      if (dis(sx + a[i].x, sy + a[i].y) < dis(sx - a[i].x, sy - a[i].y)) {
        sx += a[i].x;
        sy += a[i].y;
        ans[a[i].id] = 1;
      } else {
        sx -= a[i].x;
        sy -= a[i].y;
        ans[a[i].id] = -1;
      }
    }
  }
  for (register int i = 1, iend = n; i <= iend; ++i) printf("%d ", ans[i]);
  return 0;
}