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;
int n;
double dis(int x, int y) { return sqrt(1LL * x * x + 1LL * y * y); }
struct vec {
  int x, y;
  int num;
};
vector<vec> V;
int w[200005];
int ans[100005];
int ch[200005][2];
int cnt;
void find() {
  int s, t, q;
  for (int i = 0; i < 3; i++) {
    for (int j = i + 1; j < 3; j++) {
      if (dis(V[i].x + V[j].x, V[i].y + V[j].y) <= (1e6)) {
        s = i, t = j, q = 1;
      }
      if (dis(V[i].x - V[j].x, V[i].y - V[j].y) <= (1e6)) {
        s = i, t = j, q = -1;
      }
    }
  }
  vec p;
  p.x = V[s].x + V[t].x * q, p.y = V[s].y + V[t].y * q;
  p.num = ++cnt;
  ch[cnt][0] = V[s].num, ch[cnt][1] = V[t].num;
  w[cnt] = q;
  vector<vec>::iterator it = V.begin() + s;
  V.erase(it);
  it = V.begin() + t + (t < s ? 0 : -1);
  V.erase(it);
  V.push_back(p);
  return;
}
void dfs(int u, int t) {
  if (u <= n) {
    ans[u] = t;
    return;
  }
  dfs(ch[u][0], t);
  dfs(ch[u][1], t * w[u]);
  return;
}
int main() {
  scanf("%d", &n);
  cnt = n;
  int a = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    V.push_back((vec){x, y, i});
    if (V.size() == 3) find();
  }
  if (V.size() > 1) {
    cnt++;
    ch[cnt][0] = V[0].num;
    ch[cnt][1] = V[1].num;
    if (dis(V[0].x + V[1].x, V[0].y + V[1].y) <= (1e6 + 5e5)) {
      w[cnt] = 1;
    } else
      w[cnt] = -1;
  }
  dfs(cnt, 1);
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}