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 
#include <bits/stdc++.h>
using namespace std;
struct LLI {
 public:
  int x;
  int y;
  int inx;
  LLI(int X = 0, int Y = 0, int INX = 0) {
    x = X;
    y = Y;
    inx = INX;
  }
};
int n, i, j;
LLI now;
LLI p[100001];
bool out[100001];
bool OUT[100001];
inline long long dist(LLI a) {
  long long X = ((long long)(a.x));
  long long Y = ((long long)(a.y));
  return ((X * X) + (Y * Y));
}
inline LLI Plus(LLI a, LLI b) { return LLI(a.x + b.x, a.y + b.y); }
inline LLI Minus(LLI a, LLI b) { return LLI(a.x - b.x, a.y - b.y); }
int main() {
  scanf("%d", &n);
  for (i = 0; i != n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
    p[i].inx = i;
    if (p[i].x < 0 || (p[i].x == 0 && p[i].y < 0)) {
      p[i].x *= (-1);
      p[i].y *= (-1);
      out[i] = true;
    }
  }
  now = LLI(0, 0, 0);
  do {
    for (i = 0; i != n; i++) OUT[i] = out[i];
    now = LLI(0, 0, 0);
    random_shuffle(p, p + n);
    for (i = 0; i != n; i++) {
      if (dist(Plus(now, p[i])) <= dist(Minus(now, p[i])))
        now = Plus(now, p[i]);
      else {
        now = Minus(now, p[i]);
        OUT[p[i].inx] = (!OUT[p[i].inx]);
      }
    }
  } while (dist(now) > (1500000LL * 1500000LL));
  for (i = 0; i != n; i++) printf("%d ", OUT[i] == true ? 1 : (-1));
  printf("\n");
  return 0;
}
