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>
template <typename T>
T max(T x, T y) {
  return (x > y) ? x : y;
}
template <typename T>
T min(T x, T y) {
  return (x < y) ? x : y;
}
template <typename T>
bool chkmax(T &x, T y) {
  return (x >= y) ? 0 : (x = y, 1);
}
template <typename T>
bool chkmin(T &x, T y) {
  return (x >= y) ? 0 : (x = y, 1);
}
template <typename T>
T read(T &in) {
  in = 0;
  char c;
  T f = 1;
  while (!isdigit(c = getchar()))
    if (c == '-') f = -1;
  while (isdigit(c)) in = in * 10 + c - '0', c = getchar();
  return in *= f;
}
static const int MAX1 = 100000 + 11;
static const int MAX2 = 200000 + 11;
struct point {
  int x, y;
  point() { x = y = 0; }
  point(int _x, int _y) { x = _x, y = _y; }
  double len() { return sqrt(x * 1ll * x + y * 1ll * y); }
  point operator-() { return point(-x, -y); }
};
point operator+(point a, point b) { return point(a.x + b.x, a.y + b.y); }
point operator-(point a, point b) { return point(a.x - b.x, a.y - b.y); }
long long dots(point a, point b) { return a.x * 1ll * b.x + a.y * 1ll * b.y; }
double p_cos(point a, point b) {
  if (!a.len() || !b.len()) return 1;
  return dots(a, b) / (a.len() * b.len());
}
int n, m;
point p[MAX2];
int v[MAX2], lc[MAX2], rc[MAX2];
void DFS(int x) {
  if (!x) return;
  v[lc[x]] *= v[x], v[rc[x]] *= v[x];
  DFS(lc[x]), DFS(rc[x]);
}
int main() {
  m = read(n);
  for (int i = (1), ir = (n); i <= (ir); ++i) read(p[i].x), read(p[i].y);
  if (n == 1) {
    return puts("1"), 0;
  }
  int a = 1, b = 2;
  for (int i = (3), ir = (n); i <= (ir); ++i) {
    if (p_cos(p[a], p[b]) >= 0.5) {
      v[a] = 1, v[b] = -1, p[++m] = p[a] - p[b];
      lc[m] = a, rc[m] = b, a = m, b = i;
    } else if (p_cos(p[a], p[i]) >= 0.5) {
      v[a] = 1, v[i] = -1, p[++m] = p[a] - p[i];
      lc[m] = a, rc[m] = i, a = m;
    } else if (p_cos(p[b], p[i]) >= 0.5) {
      v[b] = 1, v[i] = -1, p[++m] = p[b] - p[i];
      lc[m] = b, rc[m] = i, b = m;
    } else if (p_cos(p[a], -p[b]) >= 0.5) {
      v[a] = 1, v[b] = 1, p[++m] = p[a] + p[b];
      lc[m] = a, rc[m] = b, a = m, b = i;
    } else if (p_cos(p[a], -p[i]) >= 0.5) {
      v[a] = 1, v[i] = 1, p[++m] = p[a] + p[i];
      lc[m] = a, rc[m] = i, a = m;
    } else {
      v[b] = 1, v[i] = 1, p[++m] = p[b] + p[i];
      lc[m] = b, rc[m] = i, b = m;
    }
  }
  v[++m] = 1, lc[m] = a, rc[m] = b;
  if ((p[a] + p[b]).len() <= (p[a] - p[b]).len())
    v[a] = v[b] = 1;
  else
    v[a] = 1, v[b] = -1;
  DFS(m);
  for (int i = (1), ir = (n); i <= (ir); ++i) printf("%d ", v[i]);
  putchar(10);
  return 0;
}