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 P {
  int x, y;
} a[200005];
P operator+(P a, P b) { return (P){a.x + b.x, a.y + b.y}; }
P operator-(P a, P b) { return (P){a.x - b.x, a.y - b.y}; }
long long len(P a) { return 1ll * a.x * a.x + 1ll * a.y * a.y; }
int n, b[200005], ch[200005][2];
queue<int> q;
void dfs(int x) {
  if (!ch[x][0]) return;
  b[ch[x][0]] *= b[x];
  b[ch[x][1]] *= b[x];
  dfs(ch[x][0]);
  dfs(ch[x][1]);
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    puts("1");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    scanf("%d%d", &a[i].x, &a[i].y), q.push(i), b[i] = 1;
  int tt = n;
  for (int i = 1; i <= n - 2; i++) {
    int x = q.front();
    q.pop();
    int y = q.front();
    q.pop();
    int z = q.front();
    q.pop();
    tt++;
    if (len(a[x] - a[y]) <= 1e12) {
      q.push(tt);
      q.push(z);
      a[tt] = a[x] - a[y];
      b[y] *= -1;
      b[tt] = 1;
      ch[tt][0] = x;
      ch[tt][1] = y;
    } else if (len(a[x] + a[y]) <= 1e12) {
      q.push(tt);
      q.push(z);
      a[tt] = a[x] + a[y];
      b[tt] = 1;
      ch[tt][0] = x;
      ch[tt][1] = y;
    } else if (len(a[x] - a[z]) <= 1e12) {
      q.push(tt);
      q.push(y);
      a[tt] = a[x] - a[z];
      b[z] *= -1;
      b[tt] = 1;
      ch[tt][0] = x;
      ch[tt][1] = z;
    } else if (len(a[x] + a[z]) <= 1e12) {
      q.push(tt);
      q.push(y);
      a[tt] = a[x] + a[z];
      b[tt] = 1;
      ch[tt][0] = x;
      ch[tt][1] = z;
    } else if (len(a[y] - a[z]) <= 1e12) {
      q.push(tt);
      q.push(x);
      a[tt] = a[y] - a[z];
      b[z] *= -1;
      b[tt] = 1;
      ch[tt][0] = y;
      ch[tt][1] = z;
    } else if (len(a[y] + a[z]) <= 1e12) {
      q.push(tt);
      q.push(x);
      a[tt] = a[y] + a[z];
      b[tt] = 1;
      ch[tt][0] = y;
      ch[tt][1] = z;
    }
  }
  int x = q.front();
  q.pop();
  int y = q.front();
  q.pop();
  tt++;
  if (len(a[x] - a[y]) <= 2.25e12) {
    q.push(tt);
    a[tt] = a[x] - a[y];
    b[y] *= -1;
    b[tt] = 1;
    ch[tt][0] = x;
    ch[tt][1] = y;
  } else if (len(a[x] + a[y]) <= 2.25e12) {
    q.push(tt);
    a[tt] = a[x] + a[y];
    b[tt] = 1;
    ch[tt][0] = x;
    ch[tt][1] = y;
  }
  dfs(tt);
  for (int i = 1; i <= n; i++) printf("%d ", b[i]);
  return 0;
}