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 int N = 2e5 + 15;
const long long MAXI = 1.5 * 1.5 * 1e12, M = 1e12;
vector<int> adj[N];
int ret[N];
long long sq(int x) { return (long long)x * x; }
long long dist(int x, int y) { return sq(x) + sq(y); }
int g[N];
int X[N], Y[N], CUR[N];
bool isgood(int a, int b, int bit, long long LIM, int pos) {
  long long u = X[a] + X[b] * bit, v = Y[a] + Y[b] * bit;
  if (dist(u, v) <= LIM) {
    X[pos] = u;
    Y[pos] = v;
    return 1;
  }
  return 0;
}
void dfs(int root, int p, int m) {
  CUR[root] *= m;
  for (auto ch : adj[root]) {
    if (ch == p) continue;
    dfs(ch, root, CUR[root]);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int x, y;
  for (int i = 1; i <= n; ++i) ret[i] = 1;
  for (int i = 0; i < N; ++i) CUR[i] = 1;
  int pos = n + 1;
  long long a = 0, b = 0, u, v;
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  int ind = 0;
  for (int i = 0; i < n; ++i) {
    g[ind++] = i;
    scanf("%d%d", &X[i], &Y[i]);
    if (ind == 3) {
      bool good = 0;
      for (int j = 0; j < 3 && !good; ++j)
        for (int k = j + 1; k < 3; ++k) {
          if (isgood(g[j], g[k], 1, M, pos)) {
            good = 1;
          } else if (isgood(g[j], g[k], -1, M, pos)) {
            good = 1;
            CUR[g[k]] *= -1;
          }
          if (good) {
            adj[pos].push_back(g[j]);
            adj[pos].push_back(g[k]);
            int rem = 3 - j - k;
            g[0] = g[rem];
            g[1] = pos++;
            ind = 2;
            break;
          }
        }
    }
  }
  adj[pos].push_back(g[0]);
  adj[pos].push_back(g[1]);
  if (!isgood(g[0], g[1], 1, MAXI, pos)) {
    assert(isgood(g[0], g[1], -1, MAXI, pos) == true);
    CUR[g[1]] *= -1;
  }
  dfs(pos, 1e9, 1);
  for (int i = 0; i < n; ++i) printf("%d ", CUR[i]);
  cout << endl;
  return 0;
}