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 MAXN = 110000;
const double LIM = 1.5E6;
const double PI = acos(-1);
struct Vector {
  int x, y;
  Vector() {}
  Vector(int a, int b) : x(a), y(b) {}
  Vector operator+(const Vector& rhs) const {
    return Vector(x + rhs.x, y + rhs.y);
  }
  Vector operator-() const { return Vector(-x, -y); }
  bool isneg() const { return x < 0 || x == 0 && y < 0; }
  double norm() const { return sqrt(1.0 * x * x + 1.0 * y * y); }
  int type() const {
    double theta = x != 0 ? atan(1.0 * y / x) : 0;
    if (theta > PI / 6) return 0;
    if (theta > -PI / 6) return 1;
    return 2;
  }
} a[MAXN];
vector<pair<int, int> > b[MAXN];
Vector s[MAXN];
vector<int> v[3];
int c[MAXN];
int merge(int x, int y) {
  if (b[x].size() < b[y].size()) {
    if (s[x].isneg() ^ s[y].isneg()) {
      for (pair<int, int> p : b[x]) b[y].push_back(p);
      s[y] = s[y] + s[x];
    } else {
      for (pair<int, int> p : b[x])
        b[y].push_back(make_pair(p.first, -p.second));
      s[y] = s[y] + (-s[x]);
    }
    b[x].clear();
    return y;
  } else {
    if (s[x].isneg() ^ s[y].isneg()) {
      for (pair<int, int> p : b[y]) b[x].push_back(p);
      s[x] = s[x] + s[y];
    } else {
      for (pair<int, int> p : b[y])
        b[x].push_back(make_pair(p.first, -p.second));
      s[x] = s[x] + (-s[y]);
    }
    b[y].clear();
    return x;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i] = Vector(x, y);
  }
  for (int i = 1; i <= n; i++) {
    b[i].push_back(make_pair(i, 1)), s[i] = a[i];
    v[a[i].type()].push_back(i);
  }
  bool flag;
  do {
    flag = 0;
    for (int i = 0; i < 3; i++)
      if (v[i].size() >= 2) {
        int x = v[i].back();
        v[i].pop_back();
        int y = v[i].back();
        v[i].pop_back();
        int z = merge(x, y);
        v[s[z].type()].push_back(z);
        flag = 1;
        break;
      }
  } while (flag);
  for (int t = 0; t < 8; t++) {
    Vector r(0, 0);
    for (int i = 0; i < 3; i++)
      if (!v[i].empty())
        if (t & 1 << i)
          r = r + s[v[i][0]];
        else
          r = r + (-s[v[i][0]]);
    if (r.norm() <= 1.5E6) {
      for (int i = 0; i < 3; i++)
        if (!v[i].empty())
          for (pair<int, int> p : b[v[i][0]])
            c[p.first] = t & 1 << i ? p.second : -p.second;
      break;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", c[i]);
  putchar('\n');
  return 0;
}