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;
template <typename T>
bool readInteger(T& x) {
  char c, r = 0, n = 0;
  x = 0;
  while (true) {
    c = getchar();
    if (c < 0 && !r)
      return 0;
    else if (c == '-' && !r)
      n = 1;
    else if (c >= '0' && c <= '9')
      x = x * 10 + c - '0', r = 1;
    else if (r)
      break;
  }
  if (n) x = -x;
  return 1;
}
template <typename T>
vector<T> readIntegral(int n) {
  vector<T> ret(n);
  for (int i = 0; i < n; i++) readInteger(ret[i]);
  return ret;
}
auto readInts = readIntegral<int>;
auto readLongs = readIntegral<long long>;
template <typename T>
vector<vector<T>> make2d(size_t r, size_t c) {
  return vector<vector<T>>(r, vector<T>(c));
}
template <typename T>
vector<vector<T>> make2d(size_t r, size_t c, const T& def) {
  return vector<vector<T>>(r, vector<T>(c, def));
}
template <typename T, T MOD>
struct ModInt {
  T value;
  ModInt() : value(0) {}
  ModInt(T x) {
    x %= MOD;
    if (x < 0) x += MOD;
    value = x;
  }

 public:
  ModInt& operator+=(ModInt x) {
    value += x.value;
    if (value >= MOD) value -= MOD;
    return *this;
  }
  ModInt& operator-=(ModInt x) {
    value -= x.value;
    if (value < 0) value += MOD;
    return *this;
  }
  ModInt& operator*=(ModInt x) {
    value *= x.value;
    value %= MOD;
    return *this;
  }
  ModInt& operator/=(ModInt x) {
    x.invert();
    return *this *= x;
  }
  ModInt operator+(ModInt x) const {
    ModInt o = *this;
    o += x;
    return o;
  }
  ModInt operator-(ModInt x) const {
    ModInt o = *this;
    o -= x;
    return o;
  }
  ModInt operator*(ModInt x) const {
    ModInt o = *this;
    o *= x;
    return o;
  }
  ModInt operator/(ModInt x) const {
    ModInt o = *this;
    o /= x;
    return o;
  }
  bool operator==(ModInt x) const { return value == x.value; }
  bool operator!=(ModInt x) const { return !(*this == x); }
  ModInt operator-() const { return ModInt(0) - *this; }
  ModInt operator^(long long x) const {
    ModInt ret = 1 % MOD;
    ModInt mul = *this;
    while (x) {
      if (x & 1) ret *= mul;
      mul *= mul;
      x >>= 1;
    }
    return ret;
  }
  ModInt& operator^=(long long x) { return *this = *this ^ x; }

 private:
  void invert() { *this ^= MOD - 2; }

 public:
  friend ostream& operator<<(ostream& out, const ModInt& x) {
    return out << x.value;
  }
};
template <typename T>
using v2d = vector<vector<T>>;
long long norm(pair<long long, long long> x) {
  return x.first * x.first + x.second * x.second;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<long long> x(n), y(n);
  vector<double> angle(n);
  for (int i = (0); i < int(n); ++i) {
    scanf("%lld%lld", &x[i], &y[i]);
    angle[i] = atan2(y[i], x[i]);
  }
  vector<int> indices(n);
  for (int i = (0); i < int(n); ++i) indices[i] = i;
  sort(indices.begin(), indices.end(),
       [&angle](int i, int j) { return angle[i] < angle[j]; });
  pair<long long, long long> cpos = {0, 0};
  vector<int> sign(n);
  for (int i : indices) {
    pair<long long, long long> a1(cpos.first + x[i], cpos.second + y[i]);
    pair<long long, long long> a2(cpos.first - x[i], cpos.second - y[i]);
    if (norm(a1) < norm(a2)) {
      sign[i] = 1;
      cpos = a1;
    } else {
      sign[i] = -1;
      cpos = a2;
    }
  }
  for (int i = (0); i < int(n); ++i) {
    printf("%d ", sign[i]);
  }
  printf("\n");
}