#include <bits/stdc++.h>
const long long inf = 0x3f3f3f3f3f3f3f3LL;
const long long mod = (long long)1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
const long long MAX = 1.5e6;
long long n;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
long long absolute(long long X, long long Y) { return X * X + Y * Y; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  vector<pair<pair<long long, long long>, long long>> v(n);
  long long idx = 0;
  for (auto& i : v) {
    cin >> i.first.first >> i.first.second;
    i.second = idx++;
  }
  while (1) {
    long long X = 0, Y = 0;
    vector<long long> ans(n);
    shuffle(v.begin(), v.end(), rng);
    for (long long j = 0; j < n; ++j) {
      long long x = v[j].first.first;
      long long y = v[j].first.second;
      long long idx = v[j].second;
      if (absolute(X + x, Y + y) < absolute(X - x, Y - y)) {
        X += x;
        Y += y;
        ans[idx] = 1;
      } else {
        X -= x;
        Y -= y;
        ans[idx] = -1;
      }
    }
    if (absolute(X, Y) <= MAX * MAX) {
      for (auto it : ans) cout << it << ' ';
      exit(0);
    }
  }
  assert(0);
  return 0;
}

### Prompt

In CPP, your task is to solve the following 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 

### Response

```cpp
#include <bits/stdc++.h>
const long long inf = 0x3f3f3f3f3f3f3f3LL;
const long long mod = (long long)1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
const long long MAX = 1.5e6;
long long n;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
long long absolute(long long X, long long Y) { return X * X + Y * Y; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  vector<pair<pair<long long, long long>, long long>> v(n);
  long long idx = 0;
  for (auto& i : v) {
    cin >> i.first.first >> i.first.second;
    i.second = idx++;
  }
  while (1) {
    long long X = 0, Y = 0;
    vector<long long> ans(n);
    shuffle(v.begin(), v.end(), rng);
    for (long long j = 0; j < n; ++j) {
      long long x = v[j].first.first;
      long long y = v[j].first.second;
      long long idx = v[j].second;
      if (absolute(X + x, Y + y) < absolute(X - x, Y - y)) {
        X += x;
        Y += y;
        ans[idx] = 1;
      } else {
        X -= x;
        Y -= y;
        ans[idx] = -1;
      }
    }
    if (absolute(X, Y) <= MAX * MAX) {
      for (auto it : ans) cout << it << ' ';
      exit(0);
    }
  }
  assert(0);
  return 0;
}

```