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 double eps = 1E-9;
const double Exp = 2.718281828459045;
const double Pi = 3.1415926535897932;
const int Max_Bit = 63;
const int INF = 2000000000;
const long long LINF = 1000000000000000007ll;
const int MOD = 1000000007;
const int NMAX = 400005;
const int FMAX = 1005;
const int KMAX = 6;
int n;
static pair<int, int> b[NMAX];
static int ans[NMAX];
void dfs(int v) {
  ans[b[v].first] = ans[v];
  if (b[v].second < 0)
    ans[(-1) * b[v].second] = -1 * ans[v];
  else
    ans[b[v].second] = ans[v];
  if (abs(b[v].first) > n) dfs(b[v].first);
  if (abs(b[v].second) > n) dfs(abs(b[v].second));
}
int main() {
  int k;
  static pair<int, int> a[NMAX];
  scanf("%d", &n);
  k = n + 1;
  priority_queue<int> q;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &a[i + 1].first, &a[i + 1].second);
    q.push((-1) * (i + 1));
  }
  while (q.size() > 2) {
    int t1 = (-1) * q.top();
    q.pop();
    int t2 = (-1) * q.top();
    q.pop();
    int t3 = (-1) * q.top();
    q.pop();
    int t[6];
    t[0] = sqrt(
        (a[t1].first + a[t2].first) * 1ll * (a[t1].first + a[t2].first) +
        (a[t1].second + a[t2].second) * 1ll * (a[t1].second + a[t2].second));
    t[1] = sqrt(
        (a[t1].first - a[t2].first) * 1ll * (a[t1].first - a[t2].first) +
        (a[t1].second - a[t2].second) * 1ll * (a[t1].second - a[t2].second));
    t[2] = sqrt(
        (a[t3].first + a[t2].first) * 1ll * (a[t3].first + a[t2].first) +
        (a[t3].second + a[t2].second) * 1ll * (a[t3].second + a[t2].second));
    t[3] = sqrt(
        (a[t3].first - a[t2].first) * 1ll * (a[t3].first - a[t2].first) +
        (a[t3].second - a[t2].second) * 1ll * (a[t3].second - a[t2].second));
    t[4] = sqrt(
        (a[t1].first + a[t3].first) * 1ll * (a[t1].first + a[t3].first) +
        (a[t1].second + a[t3].second) * 1ll * (a[t1].second + a[t3].second));
    t[5] = sqrt(
        (a[t1].first - a[t3].first) * 1ll * (a[t1].first - a[t3].first) +
        (a[t1].second - a[t3].second) * 1ll * (a[t1].second - a[t3].second));
    int tmp = INF;
    int o = 0;
    for (int i = 0; i < 6; ++i) {
      if (t[i] < tmp) {
        tmp = t[i];
        o = i;
      }
    }
    if (o == 0) {
      a[k] = make_pair(a[t1].first + a[t2].first, a[t1].second + a[t2].second);
      b[k] = make_pair(t1, t2);
      q.push((-1) * t3);
    }
    if (o == 1) {
      a[k] = make_pair(a[t1].first - a[t2].first, a[t1].second - a[t2].second);
      b[k] = make_pair(t1, t2 * (-1));
      q.push((-1) * t3);
    }
    if (o == 2) {
      a[k] = make_pair(a[t2].first + a[t3].first, a[t2].second + a[t3].second);
      b[k] = make_pair(t2, t3);
      q.push((-1) * t1);
    }
    if (o == 3) {
      a[k] = make_pair(a[t2].first - a[t3].first, a[t2].second - a[t3].second);
      b[k] = make_pair(t2, t3 * (-1));
      q.push((-1) * t1);
    }
    if (o == 4) {
      a[k] = make_pair(a[t1].first + a[t3].first, a[t1].second + a[t3].second);
      b[k] = make_pair(t1, t3);
      q.push((-1) * t2);
    }
    if (o == 5) {
      a[k] = make_pair(a[t1].first - a[t3].first, a[t1].second - a[t3].second);
      b[k] = make_pair(t1, t3 * (-1));
      q.push((-1) * t2);
    }
    q.push((-1) * k);
    k++;
  }
  int t1 = (-1) * q.top();
  q.pop();
  int t2 = (-1) * q.top();
  q.pop();
  if (sqrt((a[t1].first + a[t2].first) * (a[t1].first + a[t2].first) +
           (a[t1].second + a[t2].second) * (a[t1].second + a[t2].second)) <
      sqrt((a[t1].first - a[t2].first) * (a[t1].first - a[t2].first) +
           (a[t1].second - a[t2].second) * (a[t1].second - a[t2].second))) {
    b[k] = make_pair(t1, t2);
  } else {
    b[k] = make_pair(t1, t2 * (-1));
  }
  ans[k] = 1;
  dfs(k);
  for (int i = 0; i < n; ++i) {
    cout << ans[i + 1] << " ";
  }
  printf("\n");
}