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 mod = 1e9 + 7;
const int maxn = 1e5 + 5;
pair<int, int> e[maxn], u;
int a[maxn], b[maxn];
void solve(pair<int, int> x, pair<int, int> y, pair<int, int> z, int &xx,
           int &yy, int &zz) {
  long long dk = 1e18;
  for (xx = -1; xx <= 1; xx += 2) {
    for (yy = -1; yy <= 1; yy += 2) {
      for (zz = -1; zz <= 1; zz += 2) {
        pair<int, int> c =
            pair<int, int>(x.first * xx + y.first * yy + z.first * zz,
                           x.second * xx + y.second * yy + z.second * zz);
        if (1LL * c.first * c.first + 1LL * c.second * c.second <= 1e12) {
          u = c;
          return;
        } else
          dk = min(dk, 1LL * c.first * c.first + 1LL * c.second * c.second);
      }
    }
  }
  for (xx = -1; xx <= 1; xx += 2) {
    for (yy = -1; yy <= 1; yy += 2) {
      for (zz = -1; zz <= 1; zz += 2) {
        pair<int, int> c =
            pair<int, int>(x.first * xx + y.first * yy + z.first * zz,
                           x.second * xx + y.second * yy + z.second * zz);
        if (1LL * c.first * c.first + 1LL * c.second * c.second == dk) {
          u = c;
          return;
        }
      }
    }
  }
}
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d%d", &e[i].first, &e[i].second);
  u = e[1];
  a[1] = 1;
  for (i = 1; i + 2 <= n; i += 2)
    solve(u, e[i + 1], e[i + 2], b[i], a[i + 1], a[i + 2]);
  if (i < n) {
    for (b[i] = -1; b[i] <= 1; b[i] += 2) {
      for (a[n] = -1; a[n] <= 1; a[n] += 2) {
        pair<int, int> c = pair<int, int>(u.first * b[i] + e[n].first * a[n],
                                          u.second * b[i] + e[n].second * a[n]);
        if (1LL * c.first * c.first + 1LL * c.second * c.second <= 2.25 * 1e12)
          break;
      }
      if (a[n] <= 1) break;
    }
  }
  for (i = n; i > 0; i--) {
    if (b[i] == 0) b[i] = 1;
    if (i < n) b[i] *= b[i + 1];
    a[i] *= b[i];
  }
  for (i = 1; i <= n; i++) printf("%d%c", a[i], " \n"[i == n]);
  return 0;
}