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 str {
  long long x, y;
  bool isTerminal;
  long long sign;
  int l = -1, r = -1;
  int id;
  str(long long _x, long long _y, bool _isTerminal, long long _sign, int _id) {
    x = _x;
    y = _y;
    isTerminal = _isTerminal;
    sign = _sign;
    id = _id;
  }
  str() {}
};
const int MAX = 500500;
str nodes[MAX];
int V;
int get() { return V++; }
long long mX[MAX], mY[MAX];
int ans[MAX];
void dfs(int u, int sign) {
  if (nodes[u].isTerminal) {
    ans[nodes[u].id] = sign;
  } else {
    if (nodes[u].l != -1) dfs(nodes[u].l, sign);
    if (nodes[u].r != -1) dfs(nodes[u].r, sign * nodes[u].sign);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < int(n); i++) scanf("%lld %lld", mX + i, mY + i);
  long long p = 1000000ll;
  if (n == 1) {
    printf("1\n");
  } else {
    for (int i = 0; i < int(2); i++)
      nodes[get()] = str(mX[i], mY[i], true, 1, i);
    int u = 0, v = 1;
    for (int i = 2; i < n; i++) {
      int w = get();
      nodes[w] = str(mX[i], mY[i], true, 1, i);
      int vec[] = {u, v, w};
      int t = get();
      sort(vec, vec + 3);
      do {
        bool done = false;
        for (long long sign = -1; sign <= 1; sign += 2) {
          long long N = (nodes[vec[0]].x + sign * nodes[vec[1]].x) *
                            (nodes[vec[0]].x + sign * nodes[vec[1]].x) +
                        (nodes[vec[0]].y + sign * nodes[vec[1]].y) *
                            (nodes[vec[0]].y + sign * nodes[vec[1]].y);
          if (N <= p * p) {
            nodes[t] =
                str(nodes[vec[0]].x + sign * nodes[vec[1]].x,
                    nodes[vec[0]].y + sign * nodes[vec[1]].y, false, sign, -1);
            nodes[t].l = vec[0];
            nodes[t].r = vec[1];
            u = t;
            v = vec[2];
            done = true;
            break;
          }
        }
        if (done) break;
      } while (next_permutation(vec, vec + 3));
    }
    int t = get();
    nodes[t].l = u;
    nodes[t].r = v;
    if ((nodes[u].x - nodes[v].x) * (nodes[u].x - nodes[v].x) +
            (nodes[u].y - nodes[v].y) * (nodes[u].y - nodes[v].y) <=
        2ll * p * p) {
      nodes[t].sign = -1;
    } else {
      nodes[t].sign = 1;
    }
    dfs(t, 1);
    for (int i = 0; i < int(n); i++)
      printf("%d%c", ans[i], i == n - 1 ? '\n' : ' ');
  }
}
