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 LL = long long;
using namespace std;
const LL B = (LL)1e6 * (LL)1e6;
const int N = (int)1e5 + 10;
struct P {
  int x, y;
} ps[N];
struct E {
  int ch, sig;
};
int n, sig[N];
vector<E> ed[N];
void output(int u, int cs) {
  sig[u] = cs;
  for (E &e : ed[u]) {
    if (e.sig == 0)
      output(e.ch, -cs);
    else
      output(e.ch, cs);
  }
}
bool check(int i, int j, int s) {
  LL sx = ps[i].x, sy = ps[i].y;
  if (s) {
    sx += ps[j].x;
    sy += ps[j].y;
  } else {
    sx -= ps[j].x;
    sy -= ps[j].y;
  }
  return sx * sx + sy * sy <= B;
}
void merge(int i, int j, int s) {
  if (s) {
    ps[i].x += ps[j].x;
    ps[i].y += ps[j].y;
  } else {
    ps[i].x -= ps[j].x;
    ps[i].y -= ps[j].y;
  }
  ed[i].push_back({j, s});
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", &ps[i].x, &ps[i].y);
  deque<int> deq;
  for (int i = 0; i < n; i++) deq.push_back(i);
  while (deq.size() > 2) {
    int id[3] = {deq[0], deq[1], deq[2]};
    int ci = -1, cj = -1, ck = -1;
    for (int i = 0; i < 3; i++)
      for (int j = i + 1; j < 3; j++)
        for (int k = 0; k < 2; k++) {
          if (check(id[i], id[j], k)) {
            ci = i;
            cj = j;
            ck = k;
          }
        }
    merge(id[ci], id[cj], ck);
    deq.erase(deq.begin(), deq.begin() + 3);
    deq.push_back(id[3 - ci - cj]);
    deq.push_back(id[ci]);
  }
  if (deq.size() == 2) {
    int i1 = deq[0], i2 = deq[1];
    LL sx1 = ps[i1].x + ps[i2].x, sy1 = ps[i1].y + ps[i2].y;
    LL d1 = sx1 * sx1 + sy1 * sy1;
    LL sx2 = ps[i1].x - ps[i2].x, sy2 = ps[i1].y - ps[i2].y;
    LL d2 = sx2 * sx2 + sy2 * sy2;
    ed[i1].push_back({i2, (d1 <= d2) ? 1 : 0});
    deq.pop_back();
  }
  output(deq[0], 1);
  for (int i = 0; i < n; i++) printf("%d ", sig[i]);
  puts("");
}
