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>
void readi(T &x) {
  T input = 0;
  bool negative = false;
  char c = ' ';
  while (c < '-') {
    c = getchar();
  }
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  while (c >= '0') {
    input = input * 10 + (c - '0');
    c = getchar();
  }
  if (negative) {
    input = -input;
  }
  x = input;
}
template <class T>
void printi(T output) {
  if (output == 0) {
    putchar('0');
    return;
  }
  if (output < 0) {
    putchar('-');
    output = -output;
  }
  int aout[20];
  int ilen = 0;
  while (output) {
    aout[ilen] = ((output % 10));
    output /= 10;
    ilen++;
  }
  for (int i = ilen - 1; i >= 0; i--) {
    putchar(aout[i] + '0');
  }
  return;
}
template <class T>
void ckmin(T &a, T b) {
  a = min(a, b);
}
template <class T>
void ckmax(T &a, T b) {
  a = max(a, b);
}
long long randomize(long long mod) {
  return ((1ll << 30) * rand() + (1ll << 15) * rand() + rand()) % mod;
}
const long double PI = 4.0 * atan(1.0);
const long double EPS = 1e-20;
long long normalize(long long x, long long mod = 1000000007) {
  return (((x % mod) + mod) % mod);
}
int N;
int ans[100013];
vector<pair<pair<long long, long long>, int> > v;
pair<long long, long long> best;
long long arg(pair<long long, long long> a) {
  return (a.first * a.first) + (a.second * a.second);
}
bool cmp(pair<pair<long long, long long>, int> a,
         pair<pair<long long, long long>, int> b) {
  return arg(a.first) > arg(b.first);
}
bool cmp1(pair<pair<long long, long long>, int> a,
          pair<pair<long long, long long>, int> b) {
  return (abs(a.first.first) > abs(b.first.first));
}
bool cmp2(pair<pair<long long, long long>, int> a,
          pair<pair<long long, long long>, int> b) {
  return (abs(a.first.first) + abs(a.first.second) >
          abs(b.first.first) + abs(b.first.second));
}
bool cmp3(pair<pair<long long, long long>, int> a,
          pair<pair<long long, long long>, int> b) {
  return (abs(a.first.first) > abs(b.first.first));
}
void run() {
  best = make_pair(0, 0);
  for (int i = 0; i < N; i++) {
    pair<long long, long long> cur1 = make_pair(
        best.first + v[i].first.first, best.second + v[i].first.second);
    pair<long long, long long> cur2 = make_pair(
        best.first - v[i].first.first, best.second - v[i].first.second);
    if (arg(cur1) < arg(cur2)) {
      best = cur1;
      ans[v[i].second] = 1;
    } else {
      best = cur2;
      ans[v[i].second] = -1;
    }
  }
  if (arg(best) < 2250000000000ll) {
    for (int i = 0; i < N; i++) {
      if (i) cout << ' ';
      cout << ans[i];
    }
    exit(0);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  srand(time(0));
  if (fopen("cf995c.in", "r")) {
    freopen("cf995c.in", "r", stdin);
    freopen("cf995c.out", "w", stdout);
  }
  cin >> N;
  for (int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;
    v.push_back(make_pair(make_pair(x, y), i));
  }
  sort(v.begin(), v.end(), cmp);
  run();
  sort(v.begin(), v.end(), cmp1);
  run();
  sort(v.begin(), v.end(), cmp2);
  run();
  sort(v.begin(), v.end(), cmp3);
  run();
  random_shuffle(v.begin(), v.end(), randomize);
  run();
  assert(false);
  return 0;
}