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;
const int Mod = 1e9 + 7;
int dcmp(double x) {
  if (fabs(x) <= 1e-9) return 0;
  return x < 0 ? -1 : 1;
}
struct Point {
  double x, y;
  int id;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  bool operator<(const Point &a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
  Point operator-(const Point &a) const { return Point(x - a.x, y - a.y); }
  Point operator+(const Point &a) const { return Point(x + a.x, y + a.y); }
  Point operator*(const double &a) const { return Point(x * a, y * a); }
  Point operator/(const double &a) const { return Point(x / a, y / a); }
  void read() { scanf("%lf%lf", &x, &y); }
  void out() { cout << "debug: " << x << " " << y << endl; }
  bool operator==(const Point &a) const {
    return dcmp(x - a.x) == 0 && dcmp(y - a.y) == 0;
  }
};
double Dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double dis(Point a) { return sqrt(Dot(a, a)); }
double Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
int ConvexHull(Point *p, int n, Point *ch) {
  int m = 0;
  for (int i = 0; i <= n - 1; i++) {
    while (m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
    ch[m++] = p[i];
  }
  int k = m;
  for (int i = n - 2; i >= 0; i--) {
    while (m > k && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
    ch[m++] = p[i];
  }
  if (n > 1) m--;
  return m;
}
int Maxx = 1500000;
bool cmp(Point a, Point b) { return dcmp(dis(a) - dis(b)) > 0; }
bool check(Point o, Point p1) { return dcmp(dis(o + p1) - dis(o - p1)) <= 0; }
bool cmp1(Point a, Point b) { return a.id < b.id; }
int ans[100005];
Point p[100005];
int n, mk;
void dfs(int now, Point o) {
  if (mk) return;
  if (now == n) {
    if (dcmp(dis(o) - Maxx) > 0) return;
    mk = 1;
    for (int i = 0; i <= n - 1; i++) {
      if (i) printf(" ");
      printf("%d", ans[i]);
    }
    cout << endl;
    return;
  }
  ans[p[now].id] = 1;
  dfs(now + 1, o + p[now]);
  ans[p[now].id] = -1;
  dfs(now + 1, o - p[now]);
}
void solve() {
  cin >> n;
  for (int i = 0; i <= n - 1; i++) p[i].read(), p[i].id = i;
  mk = 0;
  sort(p, p + n, cmp);
  Point o(0, 0);
  for (int i = 0; i <= n - 1; i++) {
    if (check(o, p[i]))
      ans[p[i].id] = 1, o = o + p[i];
    else
      ans[p[i].id] = -1, o = o - p[i];
  }
  if (dcmp(dis(o) - Maxx) <= 0) {
    for (int i = 0; i <= n - 1; i++) {
      if (i) printf(" ");
      printf("%d", ans[i]);
    }
    cout << endl;
    return;
  }
  dfs(0, Point(0, 0));
}
int main() {
  int t = 0;
  solve();
  return 0;
}