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;
long long MOD = 1000LL * 1000 * 1000 + 7;
class DSU {
 private:
  vector<int> ids;
  vector<int> h;

 public:
  DSU(int n) {
    ids.resize(n);
    h.resize(n);
    for (int i = 0; i < n; i++) {
      ids[i] = i;
      h[i] = 1;
    }
  }
  int getRoot(int ind) {
    if (ids[ind] != ind) return getRoot(ids[ind]);
    return ind;
  }
  bool connected(int ind1, int ind2) { return getRoot(ind1) == getRoot(ind2); }
  void connect(int ind1, int ind2) {
    if (connected(ind1, ind2)) return;
    int r1 = getRoot(ind1);
    int r2 = getRoot(ind2);
    if (h[r1] > h[r2]) {
      ids[r2] = r1;
      h[r1] += h[r2];
    } else {
      ids[r1] = r2;
      h[r2] += h[r1];
    }
  }
  void showId(int n) {
    for (int i = 0; i < n; i++) {
      printf("%d ", ids[i]);
    }
    printf("\n");
  }
};
struct vect {
  long long x, y, id;
  long long dd() { return x * x + y * y; }
};
long long dd(long long x, long long y) { return x * x + y * y; }
int main() {
  int n;
  scanf("%d", &n);
  vector<vect> speed(n);
  vector<int> arr(n);
  for (int i = 0; i < n; i++) {
    long long a, b;
    scanf("%I64d %I64d", &a, &b);
    speed[i] = {a, b, i};
  }
  sort(speed.begin(), speed.end(),
       [](vect a, vect b) { return a.dd() < b.dd(); });
  long long curX = 0LL, curY = 0LL;
  for (int i = 0; i < n; i += 2) {
    long long r1 = dd(speed[i].x + curX, speed[i].y + curY);
    long long r2 = dd(-speed[i].x + curX, -speed[i].y + curY);
    if (r1 < r2) {
      curX += speed[i].x;
      curY += speed[i].y;
      arr[speed[i].id] = 1;
    } else {
      curX -= speed[i].x;
      curY -= speed[i].y;
      arr[speed[i].id] = -1;
    }
  }
  for (int i = 1; i < n; i += 2) {
    long long r1 = dd(speed[i].x + curX, speed[i].y + curY);
    long long r2 = dd(-speed[i].x + curX, -speed[i].y + curY);
    if (r1 < r2) {
      curX += speed[i].x;
      curY += speed[i].y;
      arr[speed[i].id] = 1;
    } else {
      curX -= speed[i].x;
      curY -= speed[i].y;
      arr[speed[i].id] = -1;
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", arr[i]);
  return 0;
}