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>
#pragma GCC optimize "-O3"
using namespace std;
struct point {
  long long x, y;
  point() { x = y = 0; }
  point(long long x, long long y) : x(x), y(y) {}
  point operator+(point a) { return point(x + a.x, y + a.y); }
  point operator-(point a) { return point(x - a.x, y - a.y); }
  long long len2() { return x * x + y * y; }
};
const int MAXN = 120000;
const long long MX1 = 1000000ll * 1000000ll;
const long long MX2 = 1500000ll * 1500000ll;
point v[MAXN];
int p[MAXN];
int up[MAXN];
int sz[MAXN];
int n;
int get(int x) {
  if (x == p[x]) return up[x];
  return up[x] ^ get(p[x]);
}
point tmp[MAXN];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> v[i].x >> v[i].y;
    tmp[i] = v[i];
  }
  for (int i = 0; i < n; ++i) {
    p[i] = i, sz[i] = 1;
  }
  vector<int> vv;
  for (int i = 0; i < n; ++i) {
    vv.push_back(i);
    int fl = 0;
    do {
      fl = 0;
      for (int j = 0; j < vv.size() && !fl; ++j)
        for (int k = j + 1; k < vv.size() && !fl; ++k) {
          if ((v[vv[j]] + v[vv[k]]).len2() <= MX1) {
            fl = 1;
            int a = vv[j];
            int b = vv[k];
            if (sz[a] < sz[b]) swap(a, b), swap(vv[j], vv[k]);
            p[b] = a;
            up[b] = 0;
            sz[a] += sz[b];
            v[a] = v[a] + v[b];
            vv.erase(vv.begin() + k);
            break;
          }
          if ((v[vv[j]] - v[vv[k]]).len2() <= MX1) {
            fl = 1;
            int a = vv[j];
            int b = vv[k];
            if (sz[a] < sz[b]) swap(a, b), swap(vv[j], vv[k]);
            p[b] = a;
            up[b] = 1;
            sz[a] += sz[b];
            v[a] = v[a] - v[b];
            vv.erase(vv.begin() + k);
            break;
          }
        }
    } while (fl);
  }
  int ans = 0;
  for (int i = 0; i < (1 << vv.size()); ++i) {
    point cur;
    for (int j = 0; j < vv.size(); ++j) {
      if ((i >> j) & 1)
        cur = cur + v[vv[j]];
      else
        cur = cur - v[vv[j]];
    }
    if (cur.len2() <= MX2) {
      ans = 1;
      for (int j = 0; j < vv.size(); ++j)
        if ((i >> j) & 1)
          up[vv[j]] = 0;
        else
          up[vv[j]] = 1;
      break;
    }
  }
  assert(ans);
  point cur;
  for (int i = 0; i < n; ++i) {
    int x = get(i);
    if (x == 0)
      cout << 1 << " ", cur = cur + tmp[i];
    else
      cout << -1 << " ", cur = cur - tmp[i];
  }
  cout << "\n";
  assert(cur.len2() <= MX2);
  return 0;
}