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;
struct QAQ {
  int first, second, i;
} zb[100005];
int n;
bool zt[100005];
inline double dis(long long x, long long y) { return sqrt(x * x + y * y); }
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> zb[i].first >> zb[i].second;
    zb[i].i = i;
  }
  while (1) {
    random_shuffle(zb + 1, zb + n + 1);
    memset(zt, 0, sizeof(zt));
    int sumx = 0, sumy = 0;
    for (int i = 1; i <= n; i++) {
      if (dis(sumx + zb[i].first, sumy + zb[i].second) <
          dis(sumx - zb[i].first, sumy - zb[i].second)) {
        zt[zb[i].i] = 1;
        sumx += zb[i].first;
        sumy += zb[i].second;
      } else {
        sumx -= zb[i].first;
        sumy -= zb[i].second;
      }
    }
    if (dis(sumx, sumy) <= 1500000) {
      for (int i = 1; i <= n; i++) zt[i] == 0 ? cout << "-1 " : cout << "1 ";
      return 0;
    }
  }
}
