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;
bool chkmax(int &x, int y) { return x < y ? x = y, true : false; }
bool chkmin(int &x, int y) { return x > y ? x = y, true : false; }
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct node {
  long long x, y, id, fl;
  double val;
  bool operator<(const node c) const { return val < c.val; }
} a[200005];
const double pi = acos(-1);
int n, m;
int c[100005];
pair<int, int> k[200005];
multiset<node> s;
void getans(int x, int t) {
  if (x <= m) return (void)(c[x] *= t * (a[x].fl ? -1 : 1));
  if (k[x].first < 0)
    getans(-k[x].first, -1 * t * (a[x].fl ? -1 : 1));
  else
    getans(k[x].first, t * (a[x].fl ? -1 : 1));
  if (k[x].second < 0)
    getans(-k[x].second, -1 * t * (a[x].fl ? -1 : 1));
  else
    getans(k[x].second, t * (a[x].fl ? -1 : 1));
}
int main() {
  n = m = readint();
  for (int i = 1; i <= n; i++)
    a[i].x = readint(), a[i].y = readint(), a[i].id = i;
  if (n == 1) return printf("1\n"), 0;
  for (int i = 1; i <= n; i++) c[i] = 1;
  for (int i = 1; i <= n; i++)
    if (a[i].y < 0 || (a[i].y == 0 && a[i].x < 0))
      a[i].x *= -1, a[i].y *= -1, a[i].fl = true;
  for (int i = 1; i <= n; i++) a[i].val = atan2(a[i].y, a[i].x);
  for (int i = 1; i <= n; i++) s.insert(a[i]);
  while (s.size() > 2) {
    bool fl = false;
    for (auto it = s.begin(), it1 = (++s.begin()); it1 != s.end();
         it++, it1++) {
      if (it1->val - it->val <= pi / 3) {
        k[++n] = make_pair(it->id, -it1->id);
        a[n].x = it->x - it1->x, a[n].y = it->y - it1->y, a[n].id = n;
        if (a[n].y < 0 || (a[n].y == 0 && a[n].x < 0))
          a[n].x *= -1, a[n].y *= -1, a[n].fl = true;
        a[n].val = atan2(a[n].y, a[n].x);
        s.insert(a[n]);
        s.erase(it1), s.erase(it);
        fl = true;
        break;
      }
    }
    if (!fl) {
      auto it = s.begin(), it1 = (--s.end());
      k[++n] = make_pair(it->id, it1->id);
      a[n].x = it->x + it1->x, a[n].y = it->y + it1->y, a[n].id = n;
      if (a[n].y < 0 || (a[n].y == 0 && a[n].x < 0))
        a[n].x *= -1, a[n].y *= -1, a[n].fl = true;
      a[n].val = atan2(a[n].y, a[n].x);
      s.insert(a[n]);
      s.erase(it1), s.erase(it);
      break;
    }
  }
  auto it = s.begin(), it1 = ++s.begin();
  if (it->x * it1->x + it->y * it1->y > 0) {
    k[++n] = make_pair(it->id, -it1->id);
    a[n].x = it->x - it1->x, a[n].y = it->y - it1->y, a[n].id = n;
    if (a[n].y < 0 || (a[n].y == 0 && a[n].x < 0))
      a[n].x *= -1, a[n].y *= -1, a[n].fl = true;
    a[n].val = atan2(a[n].y, a[n].x);
    s.insert(a[n]);
    s.erase(it1), s.erase(it);
  } else {
    k[++n] = make_pair(it->id, it1->id);
    a[n].x = it->x + it1->x, a[n].y = it->y + it1->y, a[n].id = n;
    if (a[n].y < 0 || (a[n].y == 0 && a[n].x < 0))
      a[n].x *= -1, a[n].y *= -1, a[n].fl = true;
    a[n].val = atan2(a[n].y, a[n].x);
    s.insert(a[n]);
    s.erase(it1), s.erase(it);
  }
  getans(n, 1);
  for (int i = 1; i <= m; i++) printf("%d ", c[i]);
  printf("\n");
  return 0;
}