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;
using lint = long long;
using pi = complex<int>;
const int mod = 1e9 + 7;
const int MAXN = 200005;
int n;
pi a[MAXN];
int l[MAXN], r[MAXN], ret[MAXN];
lint sz(pi a) { return norm(complex<lint>(a)); }
void dfs(int x, int v) {
  if (x < n) {
    ret[x] = v;
    return;
  }
  if (a[l[x]] - a[r[x]] == a[x]) {
    dfs(l[x], v);
    dfs(r[x], -v);
  } else {
    dfs(l[x], v);
    dfs(r[x], v);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int re, im;
    scanf("%d %d", &re, &im);
    a[i] = {re, im};
  };
  if (n == 1) {
    puts("1");
    return 0;
  }
  int p1 = 0, p2 = 1, piv = n;
  for (int i = 2; i < n; i++) {
    lint rad = max({sz(a[p1]), sz(a[p2]), sz(a[i])});
    if (sz(a[p1] + a[i]) <= rad) {
      l[piv] = p1;
      r[piv] = i;
      a[piv] = a[p1] + a[i];
      tie(p1, p2) = {piv, p2};
    } else if (sz(a[p1] - a[i]) <= rad) {
      l[piv] = p1;
      r[piv] = i;
      a[piv] = a[p1] - a[i];
      tie(p1, p2) = {piv, p2};
    } else if (sz(a[p2] + a[i]) <= rad) {
      l[piv] = p2;
      r[piv] = i;
      a[piv] = a[p2] + a[i];
      tie(p1, p2) = {p1, piv};
    } else if (sz(a[p2] - a[i]) <= rad) {
      l[piv] = p2;
      r[piv] = i;
      a[piv] = a[p2] - a[i];
      tie(p1, p2) = {p1, piv};
    } else if (sz(a[p1] + a[p2]) <= rad) {
      l[piv] = p1;
      r[piv] = p2;
      a[piv] = a[p1] + a[p2];
      tie(p1, p2) = {piv, i};
    } else if (sz(a[p1] - a[p2]) <= rad) {
      l[piv] = p1;
      r[piv] = p2;
      a[piv] = a[p1] - a[p2];
      tie(p1, p2) = {piv, i};
    } else
      assert(0);
    piv++;
  }
  if (sz(a[p1] + a[p2]) < sz(a[p1] - a[p2])) {
    dfs(p1, 1);
    dfs(p2, 1);
  } else {
    dfs(p1, 1);
    dfs(p2, -1);
  }
  for (int i = 0; i < n; i++) printf("%d ", ret[i]);
}