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;
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
int n;
int ans[100001];
const long long inf = 2250000000000;
inline long long cal(const pair<pair<long long, long long>, long long> &first) {
  return first.first.first * first.first.first +
         first.first.second * first.first.second;
}
inline long long mul(long long first, long long second) {
  return first * first + second * second;
}
inline bool cmp(const pair<pair<long long, long long>, long long> &a,
                const pair<pair<long long, long long>, long long> &b) {
  return cal(a) < cal(b);
}
pair<pair<long long, long long>, long long> a[100001];
void enter() {
  read(n);
  for (int i = 0; i < n; i++) {
    read(a[i].first.first);
    read(a[i].first.second);
    a[i].second = i;
  }
  sort(a, a + n, cmp);
}
void add(long long &first, long long second) { first = first + second; }
void sub(long long &first, long long second) { first = first - second; }
void progress() {
  long long first = 0, second = 0;
  long long xx = 0, yy = 0;
  for (int i = 0; i < n; i++) {
    add(xx, a[i].first.first);
    add(yy, a[i].first.second);
    if (mul(first + a[i].first.first, second + a[i].first.second) <=
        mul(first, second)) {
      add(second, a[i].first.second);
      add(first, a[i].first.first);
      ans[a[i].second] = 1;
    } else {
      sub(second, a[i].first.second);
      sub(first, a[i].first.first);
      ans[a[i].second] = -1;
    }
  }
  while (mul(first, second) > inf) {
    for (int i = n - 1; i >= 0; i--) {
      if (ans[a[i].second] == -1 &&
          mul(first + 2 * a[i].first.first, second + 2 * a[i].first.second) <=
              inf) {
        add(second, 2 * a[i].first.second);
        add(first, 2 * a[i].first.first);
        ans[a[i].second] = 1;
      } else if (ans[a[i].second] == 1 &&
                 mul(first - 2 * a[i].first.first,
                     second - 2 * a[i].first.second) <= inf) {
        sub(first, 2 * a[i].first.first);
        sub(second, 2 * a[i].first.second);
        ans[a[i].second] = -1;
      }
    }
  }
}
int main() {
  enter();
  progress();
  for (int i = 0; i < n; i++) {
    write(ans[i]);
    putchar(' ');
  }
}