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 maxn = 150005;
long long up;
int n, m, i, j, k;
struct sb {
  int x, y, z;
};
sb a[maxn];
int ans[maxn];
inline int get() {
  char c;
  while (((c = getchar()) < 48 || c > 57) && c != '-')
    ;
  if (c == '-') {
    int res = 0;
    while ((c = getchar()) >= 48 && c <= 57) res = res * 10 + c - '0';
    return -res;
  } else {
    int res = c - '0';
    while ((c = getchar()) >= 48 && c <= 57) res = res * 10 + c - '0';
    return res;
  }
}
inline bool check(int x, int y) { return (1LL * x * x + 1LL * y * y <= up); }
inline int dfs(int k, int x, int y) {
  if (k > n) {
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
    exit(0);
  }
  if (check(x + a[k].x, y + a[k].y))
    ans[a[k].z] = 1, dfs(k + 1, a[k].x + x, a[k].y + y);
  if (check(x - a[k].x, y - a[k].y))
    ans[a[k].z] = -1, dfs(k + 1, x - a[k].x, y - a[k].y);
}
int main() {
  cin >> n;
  up = 1500000;
  up = up * up;
  for (i = 1; i <= n; i++) a[i].x = get(), a[i].y = get(), a[i].z = i;
  random_shuffle(a + 1, a + 1 + n);
  dfs(1, 0, 0);
}