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 read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
__inline string next_line() {
  const int SIZE = 1100000;
  static char buf[SIZE + 1];
  static int p = SIZE;
  string res;
  int f = 1;
  while ((p < SIZE || (p = 0, buf[fread(buf, 1, SIZE, stdin)] = 0, buf[0])) &&
         (buf[p] != '\n' && (res += buf[p], f = 0, 1) || f))
    ++p;
  return res;
}
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 1e5 + 5;
const int MOD = 1e9 + 7;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
const long long MAX = 225 * 1e10;
int n, ans[MAXN];
struct Point {
  long long x, y;
  int id;
  long long len() { return x * x + y * y; }
  Point() {}
  Point(long long _x, long long _y) {
    x = _x;
    y = _y;
  }
  Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
  Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
} p, a[MAXN], t1, t2;
bool cmp(Point a, Point b) { return a.len() > b.len(); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d%I64d", &a[i].x, &a[i].y);
    a[i].id = i;
  }
  while (1) {
    p.x = p.y = 0;
    random_shuffle(a + 1, a + 1 + n);
    for (int i = 1; i <= n; i++) {
      t1 = p + a[i];
      t2 = p - a[i];
      if (t1.len() < t2.len())
        p = t1, ans[a[i].id] = 1;
      else
        p = t2, ans[a[i].id] = -1;
    }
    if (p.len() <= MAX) {
      for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
      break;
    }
  }
}