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;
long long mod = 1000000007;
long long mod2 = 998244353;
long long mod3 = 1000003;
long long mod4 = 998244853;
long long mod5 = 1000000009;
long long inf = 1LL << 62;
int iinf = 1 << 30;
double pi = 3.141592653589793238462643383279L;
double eps = 1e-14;
int dh[4] = {1, -1, 0, 0};
int dw[4] = {0, 0, 1, -1};
int ddh[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddw[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long c = gcd(a, b);
  return a * b / c;
}
long long Pow(long long n, long long k) {
  if (k < 0) return 0;
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k & 1) ret *= now;
    now *= now;
    k /= 2;
  }
  return ret;
}
long long beki(long long n, long long k, long long md) {
  long long ret = 1;
  long long now = n;
  now %= md;
  while (k > 0) {
    if (k % 2 == 1) {
      ret *= now;
      ret %= md;
    }
    now *= now;
    now %= md;
    k /= 2;
  }
  return ret;
}
long long gyaku(long long n, long long md) { return beki(n, md - 2, md); }
long long poc(long long n) {
  long long ret = 0;
  long long u = n;
  while (u > 0) {
    ret += u % 2;
    u /= 2;
  }
  return ret;
}
double x[100010], y[100010];
long long num[100010];
double si(double x, double y) {
  if (x == 0) {
    if (y >= 0) {
      return pi / 2;
    } else
      return 3 * pi / 2;
  } else {
    double ret = atan(y / x);
    if (x <= 0 && y >= 0)
      ret += pi;
    else if (x < 0 && y < 0)
      ret += pi;
    else if (x > 0 && y < 0)
      ret += 2 * pi;
    return ret;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  double nx = 0, ny = 0;
  set<pair<double, int>> st;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    nx -= x[i];
    ny -= y[i];
    st.insert({si(x[i], y[i]), i});
    num[i] = -1;
  }
  for (;;) {
    if (nx * nx + ny * ny <= 2.25 * 1000000000000) break;
    double sii = si(-nx, -ny);
    pair<double, int> p = {sii, -1};
    auto itr = st.lower_bound(p);
    if (itr != st.end()) {
      pair<double, int> pp = *itr;
      int nu = pp.second;
      if (itr != st.begin()) {
        itr--;
        pp = *itr;
        int nuu = pp.second;
        double nnx = nx + 2 * x[nuu], nny = ny + 2 * y[nuu];
        double nnxx = nx + 2 * x[nu], nnyy = ny + 2 * y[nu];
        if (nnx * nnx + nny * nny <= nnxx * nnxx + nnyy * nnyy) {
          num[nuu] = 1;
          nx = nnx;
          ny = nny;
          st.erase(itr);
        } else {
          nx = nnxx;
          ny = nnyy;
          num[nu] = 1;
          itr++;
          st.erase(itr);
        }
      } else {
        double nnxx = nx + 2 * x[nu], nnyy = ny + 2 * y[nu];
        nx = nnxx;
        ny = nnyy;
        num[nu] = 1;
        st.erase(itr);
      }
    } else {
      itr--;
      pair<double, int> pp = *itr;
      int nu = pp.second;
      double nnxx = nx + 2 * x[nu], nnyy = ny + 2 * y[nu];
      nx = nnxx;
      ny = nnyy;
      num[nu] = 1;
      st.erase(itr);
    }
  }
  for (int i = 0; i < n; i++) cout << num[i] << " ";
  cout << endl;
}