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;
static inline void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
static inline int ctz(unsigned x) { return __builtin_ctz(x); }
static inline int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
static inline int clz(unsigned x) { return __builtin_clz(x); }
static inline int clzll(unsigned long long x) { return __builtin_clzll(x); }
static inline int popcnt(unsigned x) { return __builtin_popcount(x); }
static inline int popcntll(unsigned long long x) {
  return __builtin_popcountll(x);
}
static inline int bsr(unsigned x) { return 31 ^ clz(x); }
static inline int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
const long long lmax = 1500000;
const long long sqrmax = lmax * lmax;
struct V {
  int x, y, i;
};
int quart(const V &a) {
  if (a.y >= 0)
    return a.x >= 0 ? 0 : 1;
  else
    return a.x >= 0 ? 3 : 2;
}
bool cmp(const V &a, const V &b) {
  if (quart(a) != quart(b)) return quart(a) < quart(b);
  return (long long)a.x * b.y > (long long)a.y * b.x;
}
long long len(long long x, long long y) {
  if (abs(x) > lmax || abs(y) > lmax) return 2 * sqrmax;
  return x * x + y * y;
}
V v[100016];
int c[100016];
bool dfs(int n, int i = 0, long long x = 0, long long y = 0) {
  if (i == n) return len(x, y) <= sqrmax;
  pair<long long, int> w[2];
  w[0] = {len(x + v[i].x, y + v[i].y), 1};
  w[1] = {len(x - v[i].x, y - v[i].y), -1};
  if (w[0].first > w[1].first) swap(w[0], w[1]);
  for (int j = 0; j < 2; ++j) {
    int sig = w[j].second;
    c[v[i].i] = sig;
    if (dfs(n, i + 1, x + sig * v[i].x, y + sig * v[i].y)) return true;
  }
  return false;
}
int main() {
  canhazfast();
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> v[i].x >> v[i].y, v[i].i = i;
  sort(v, v + n, cmp);
  dfs(n);
  for (int i = 0; i < n; ++i) cout << c[i] << ' ';
  return 0;
}