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 maxn = 1e5 + 100;
const int mod = 1e9 + 7;
using ll = long long;
struct Point {
  int x, y, id;
  Point operator*(const int &a) const { return Point{x * a, y * a}; }
  Point operator+(const Point &a) const { return Point{x + a.x, y + a.y}; }
} a[maxn];
const ll M = 1.5e6 * 1.5e6;
inline ll getdis(Point a) { return 1LL * a.x * a.x + 1LL * a.y * a.y; }
bool cmp(const Point &a, const Point &b) { return a.x * b.y < a.y * b.x; }
inline Point check(Point &a, Point &b) {
  ll dis;
  dis = getdis(a + b);
  if (dis <= M) return {1, 1};
  dis = getdis(a + (b * -1));
  if (dis <= M) return {1, -1};
  dis = getdis((a * -1) + b);
  if (dis <= M) return {-1, 1};
  dis = getdis((a * -1) + (b * -1));
  if (dis <= M) return {-1, -1};
  return {0, 0};
}
int ans[maxn], lazy[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y), a[i].id = i;
  for (int i = 1; i <= n; ++i) ans[i] = 1, lazy[i] = 1;
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i < n; ++i) {
    Point c = {0, 0};
    for (int j = i + 1; j <= n; ++j) {
      c = check(a[i], a[j]);
      if (c.x != 0) {
        swap(a[j], a[i + 1]);
        break;
      }
    }
    if (c.x == 0) c = {1, 1};
    ans[a[i + 1].id] = c.y;
    a[i + 1] = a[i] * c.x + a[i + 1] * c.y;
    lazy[i] = c.x;
  }
  int sta = 1;
  for (int i = n; i >= 1; --i) {
    sta *= lazy[i];
    ans[a[i].id] *= sta;
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}