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 <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 400001;
pair<int, int> a[maxn], b[maxn];
int s[maxn];
int rev[maxn], ans[maxn];
int n;
inline pair<int, int> add(const pair<int, int> &lhs, const pair<int, int> &rhs,
                          const int inv) {
  return pair<int, int>(lhs.first + inv * rhs.first,
                        lhs.second + inv * rhs.second);
}
inline long long len(const pair<int, int> &x) {
  return 1ll * x.first * x.first + 1ll * x.second * x.second;
}
pair<int, int> gao(int l, int r) {
  if (r == l) return a[l];
  if (l + 1 == r) {
    int mid = (l + r) >> 1;
    auto x = gao(l, mid);
    auto y = gao(mid + 1, r);
    int dx = -1, dy = -1;
    pair<int, int> now = pair<int, int>(0, 0);
    pair<int, int> tmp;
    for (int i : {-1, 1})
      for (int j : {-1, 1}) {
        auto fin = add(now, x, i);
        fin = add(fin, y, j);
        if (i + j == -2)
          tmp = fin;
        else if (len(tmp) > len(fin)) {
          dx = i, dy = j;
          tmp = fin;
        }
      }
    if (dx == -1) rev[l]++, rev[mid + 1]--;
    if (dy == -1) rev[mid + 1]++, rev[r + 1]--;
    return tmp;
  } else {
    int l1 = l + (r - l + 1) / 3 - 1;
    int l2 = (l1 + 1 + r) >> 1;
    auto x = gao(l, l1);
    auto y = gao(l1 + 1, l2);
    auto z = gao(l2 + 1, r);
    int dx = -1, dy = -1, dz = -1;
    pair<int, int> now = pair<int, int>(0, 0);
    pair<int, int> tmp;
    for (int i : {-1, 1})
      for (int j : {-1, 1})
        for (int k : {-1, 1}) {
          auto fin = add(now, x, i);
          fin = add(fin, y, j);
          fin = add(fin, z, k);
          if (i + j + k == -3)
            tmp = fin;
          else if (len(tmp) > len(fin)) {
            dx = i, dy = j, dz = k;
            tmp = fin;
          }
        }
    if (dx == -1) rev[l]++, rev[l1 + 1]--;
    if (dy == -1) rev[l1 + 1]++, rev[l2 + 1]--;
    if (dz == -1) rev[l2 + 1]++, rev[r + 1]--;
    return tmp;
  }
}
int main() {
  srand(time(NULL));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &b[i].first, &b[i].second);
  for (int i = 1; i <= n; i++) s[i] = i;
  pair<int, int> tmp;
  do {
    random_shuffle(s + 1, s + 1 + n);
    for (int i = 1; i <= n; i++) a[i] = b[s[i]];
    for (int i = 0; i <= n + 1; i++) rev[i] = 0;
    tmp = gao(1, n);
  } while (len(tmp) > 2250000000000ll);
  int now = 0;
  for (int i = 1; i <= n; i++) {
    now += rev[i];
    ans[s[i]] = (now & 1) ? -1 : 1;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}