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;
struct Node {
  int id, tag;
  vector<Node*> child;
  Node(int);
};
Node::Node(int id) {
  this->id = id;
  this->tag = 1;
}
const int MAXN = 1e5 + 10, V = 1e6;
int n, x[MAXN], y[MAXN], ans[MAXN];
Node* nod[MAXN];
long long sqr(int x) { return (long long)x * x; }
Node* merge(Node* a, Node* b) {
  Node* c = new Node(0);
  c->child.push_back(a);
  c->child.push_back(b);
  return c;
}
void dfs(Node* a, int b) {
  b *= a->tag;
  if (a->id) ans[a->id] = b;
  for (vector<Node*>::iterator i = a->child.begin(); i < a->child.end(); i++) {
    dfs(*i, b);
  }
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    nod[i] = new Node(i);
  }
  for (int m = n; m > 2; m--) {
    if (sqr(x[m - 1] + x[m]) + sqr(y[m - 1] + y[m]) <= sqr(V)) {
      x[m - 1] += x[m], y[m - 1] += y[m];
      nod[m - 1] = merge(nod[m - 1], nod[m]);
    } else if (sqr(x[m - 1] - x[m]) + sqr(y[m - 1] - y[m]) <= sqr(V)) {
      x[m - 1] -= x[m], y[m - 1] -= y[m], nod[m]->tag = -1;
      nod[m - 1] = merge(nod[m - 1], nod[m]);
    } else if (sqr(x[m - 2] + x[m]) + sqr(y[m - 2] + y[m]) <= sqr(V)) {
      x[m - 2] += x[m], y[m - 2] += y[m];
      nod[m - 2] = merge(nod[m - 2], nod[m]);
    } else if (sqr(x[m - 2] - x[m]) + sqr(y[m - 2] - y[m]) <= sqr(V)) {
      x[m - 2] -= x[m], y[m - 2] -= y[m], nod[m]->tag = -1;
      nod[m - 2] = merge(nod[m - 2], nod[m]);
    } else if (sqr(x[m - 2] + x[m - 1]) + sqr(y[m - 2] + y[m - 1]) <= sqr(V)) {
      x[m - 2] += x[m - 1], y[m - 2] += y[m - 1];
      nod[m - 2] = merge(nod[m - 2], nod[m - 1]);
      x[m - 1] = x[m], y[m - 1] = y[m], nod[m - 1] = nod[m];
    } else {
      x[m - 2] -= x[m - 1], y[m - 2] -= y[m - 1], nod[m - 1]->tag = -1;
      nod[m - 2] = merge(nod[m - 2], nod[m - 1]);
      x[m - 1] = x[m], y[m - 1] = y[m], nod[m - 1] = nod[m];
    }
  }
  if (sqr(x[1] + x[2]) + sqr(y[1] + y[2]) <
      sqr(x[1] - x[2]) + sqr(y[1] - y[2])) {
    x[1] += x[2], y[1] += y[2];
    nod[1] = merge(nod[1], nod[2]);
  } else {
    x[1] -= x[2], y[1] -= y[2], nod[2]->tag = -1;
    nod[1] = merge(nod[1], nod[2]);
  }
  dfs(nod[1], 1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
}