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 
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
const int R = 1000000;
const double eps = 1e-8;
struct P {
  double x, y;
  P() {}
  P(double xx, double yy) : x(xx), y(yy) {}
  P operator+(P rhs) { return P(x + rhs.x, y + rhs.y); }
  P operator-(P rhs) { return P(x - rhs.x, y - rhs.y); }
  P operator*(double k) { return P(k * x, k * y); }
  double dot(P rhs) { return x * rhs.x + y * rhs.y; }
  double det(P rhs) { return x * rhs.y - y * rhs.x; }
};
struct node {
  int ls, rs, type;
};
int n, tot = 0;
P vec[2 * N];
node nd[2 * N];
int ans[N];
void dfs(int id, int type) {
  if (nd[id].ls == -1) {
    ans[id] = type;
    return;
  }
  dfs(nd[id].ls, type);
  dfs(nd[id].rs, type * nd[id].type);
}
int main() {
  scanf("%d", &n);
  queue<int> q;
  for (int i = 0; i < n; i++) {
    scanf("%lf%lf", &vec[tot].x, &vec[tot].y);
    nd[tot++] = {-1, -1, 1};
    q.push(i);
  }
  while (q.size() > 2) {
    int a[3] = {-1};
    for (int i = 0; i < 3; i++) {
      a[i] = q.front();
      q.pop();
    }
    int x = -1, y = -1, type = 1;
    for (int i = 0; i < 3 && x == -1; i++) {
      for (int j = i + 1; j < 3 && x == -1; j++) {
        if (sqrt((vec[a[i]] + vec[a[j]]).dot(vec[a[i]] + vec[a[j]])) <
            R + eps) {
          x = a[i], y = a[j];
          break;
        }
        if (sqrt((vec[a[i]] - vec[a[j]]).dot(vec[a[i]] - vec[a[j]])) <
            R + eps) {
          x = a[i], y = a[j], type = -1;
          break;
        }
      }
    }
    for (int i = 0; i < 3; i++) {
      if (a[i] != x && a[i] != y) q.push(a[i]);
    }
    nd[tot] = {x, y, type};
    vec[tot] = vec[x] + (vec[y] * type);
    q.push(tot++);
  }
  if (q.size() < 2) {
    printf("1\n");
    return 0;
  }
  int x = q.front();
  q.pop();
  int y = q.front();
  q.pop();
  if (sqrt((vec[x] + vec[y]).dot(vec[x] + vec[y])) < sqrt(2) * R + eps) {
    nd[tot++] = {x, y, 1};
  } else {
    nd[tot++] = {x, y, -1};
  }
  dfs(tot - 1, 1);
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  printf("\n");
}
