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 gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gli() {
  long long a;
  scanf("%I64d", &a);
  return a;
}
int x[200000];
int y[200000];
int w[200000];
int f[200000];
int c1[200000];
int c2[200000];
int L;
int mknd(int ww) {
  x[L] = gi();
  y[L] = gi();
  w[L] = ww;
  f[L] = 1;
  return L++;
}
int mkp(int n1, int n2, int m) {
  f[n2] = m;
  x[L] = x[n1] + m * x[n2];
  y[L] = y[n1] + m * y[n2];
  f[L] = 1;
  c1[L] = n1;
  c2[L] = n2;
  return L++;
}
void st(int a, int ff) {
  ff *= f[a];
  if (w[a] == -1) {
    st(c1[a], ff);
    st(c2[a], ff);
  } else
    x[w[a]] = ff;
}
int sm(long long x, long long y, long long c = 1000000LL) {
  return x * x + y * y <= c * c;
}
int main() {
  int n = gi();
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  memset(w, -1, sizeof(w));
  L = 0;
  int n1 = mknd(0);
  int n2 = mknd(1);
  for (int i = 2; i < n; i++) {
    int n3 = mknd(i);
    if (sm(x[n1] + x[n2], y[n1] + y[n2])) {
      n1 = mkp(n1, n2, 1);
      n2 = n3;
    } else if (sm(x[n1] + x[n3], y[n1] + y[n3])) {
      n1 = mkp(n1, n3, 1);
    } else if (sm(x[n2] + x[n3], y[n2] + y[n3])) {
      n2 = mkp(n2, n3, 1);
    } else if (sm(x[n1] - x[n2], y[n1] - y[n2])) {
      n1 = mkp(n1, n2, -1);
      n2 = n3;
    } else if (sm(x[n1] - x[n3], y[n1] - y[n3])) {
      n1 = mkp(n1, n3, -1);
    } else {
      n2 = mkp(n2, n3, -1);
    }
  }
  st(mkp(n1, n2, sm(x[n1] + x[n2], y[n1] + y[n2], 1500000LL) ? 1 : -1), 1);
  for (int i = 0; i < n; i++) printf("%d%c", x[i], (i == n - 1 ? '\n' : ' '));
  return 0;
}