#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
pair<long long, long long> add(pair<long long, long long> p1,
                               pair<long long, long long> p2) {
  return pair<long long, long long>(p1.first + p2.first, p1.second + p2.second);
}
long long dist(pair<long long, long long> p) {
  return p.first * p.first + p.second * p.second;
}
pair<long long, long long> operator+(const pair<long long, long long> &p1,
                                     const pair<long long, long long> &p2) {
  return pair<long long, long long>(p1.first + p2.first, p1.second + p2.second);
}
pair<long long, long long> operator-(const pair<long long, long long> &p1,
                                     const pair<long long, long long> &p2) {
  return pair<long long, long long>(p1.first - p2.first, p1.second - p2.second);
}
pair<long long, long long> operator-(const pair<long long, long long> &p) {
  return pair<long long, long long>(-p.first, -p.second);
}
int N;
pair<long long, long long> moves[MAXN];
pair<int, int> below[2 * MAXN];
int inv[2 * MAXN];
bool out[MAXN];
void dfs(int pos, bool has_inv) {
  has_inv ^= inv[pos];
  if (pos <= N) {
    out[pos] = has_inv;
    return;
  }
  dfs(below[pos].first, has_inv);
  dfs(below[pos].second, has_inv);
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  srand(time(0));
  cin >> N;
  for (int a = 1; a <= N; a++) {
    cin >> moves[a].first >> moves[a].second;
  }
  pair<long long, long long> vec1 = moves[1], vec2 = moves[2];
  int id1 = 1, id2 = 2;
  for (int a = 3; a <= N; a++) {
    pair<long long, long long> vec3 = moves[a];
    int id3 = a;
    int new_id = a + N - 2;
    pair<long long, long long> res11 = vec1 + vec2;
    pair<long long, long long> res12 = vec1 - vec2;
    pair<long long, long long> res21 = vec1 + vec3;
    pair<long long, long long> res22 = vec1 - vec3;
    pair<long long, long long> res31 = vec2 + vec3;
    pair<long long, long long> res32 = vec2 - vec3;
    long long min_dist = 1e18;
    min_dist = min(min_dist, dist(res11));
    min_dist = min(min_dist, dist(res12));
    min_dist = min(min_dist, dist(res21));
    min_dist = min(min_dist, dist(res22));
    min_dist = min(min_dist, dist(res31));
    min_dist = min(min_dist, dist(res32));
    if (dist(res11) == min_dist) {
      below[new_id] = pair<int, int>(id1, id2);
      id1 = new_id;
      id2 = id3;
      vec1 = res11;
      vec2 = vec3;
    } else if (dist(res12) == min_dist) {
      below[new_id] = pair<int, int>(id1, id2);
      inv[id2] = true;
      id1 = new_id;
      id2 = id3;
      vec1 = res12;
      vec2 = vec3;
    } else if (dist(res21) == min_dist) {
      below[new_id] = pair<int, int>(id1, id3);
      id1 = new_id;
      vec1 = res21;
    } else if (dist(res22) == min_dist) {
      below[new_id] = pair<int, int>(id1, id3);
      inv[id3] = true;
      id1 = new_id;
      vec1 = res22;
    } else if (dist(res31) == min_dist) {
      below[new_id] = pair<int, int>(id2, id3);
      id2 = new_id;
      vec2 = res31;
    } else if (dist(res32) == min_dist) {
      below[new_id] = pair<int, int>(id2, id3);
      inv[id3] = true;
      id2 = new_id;
      vec2 = res32;
    } else {
      assert(false);
    }
  }
  dfs(id1, false);
  if (dist(vec1 - vec2) <= dist(vec1 + vec2)) {
    inv[id2] = true;
  }
  dfs(id2, false);
  for (int a = 1; a <= N; a++) {
    if (out[a]) {
      cout << "-1 ";
    } else {
      cout << "1 ";
    }
  }
}

### Prompt

Please create a solution in Cpp to 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>
using namespace std;
const int MAXN = 1e5 + 5;
pair<long long, long long> add(pair<long long, long long> p1,
                               pair<long long, long long> p2) {
  return pair<long long, long long>(p1.first + p2.first, p1.second + p2.second);
}
long long dist(pair<long long, long long> p) {
  return p.first * p.first + p.second * p.second;
}
pair<long long, long long> operator+(const pair<long long, long long> &p1,
                                     const pair<long long, long long> &p2) {
  return pair<long long, long long>(p1.first + p2.first, p1.second + p2.second);
}
pair<long long, long long> operator-(const pair<long long, long long> &p1,
                                     const pair<long long, long long> &p2) {
  return pair<long long, long long>(p1.first - p2.first, p1.second - p2.second);
}
pair<long long, long long> operator-(const pair<long long, long long> &p) {
  return pair<long long, long long>(-p.first, -p.second);
}
int N;
pair<long long, long long> moves[MAXN];
pair<int, int> below[2 * MAXN];
int inv[2 * MAXN];
bool out[MAXN];
void dfs(int pos, bool has_inv) {
  has_inv ^= inv[pos];
  if (pos <= N) {
    out[pos] = has_inv;
    return;
  }
  dfs(below[pos].first, has_inv);
  dfs(below[pos].second, has_inv);
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  srand(time(0));
  cin >> N;
  for (int a = 1; a <= N; a++) {
    cin >> moves[a].first >> moves[a].second;
  }
  pair<long long, long long> vec1 = moves[1], vec2 = moves[2];
  int id1 = 1, id2 = 2;
  for (int a = 3; a <= N; a++) {
    pair<long long, long long> vec3 = moves[a];
    int id3 = a;
    int new_id = a + N - 2;
    pair<long long, long long> res11 = vec1 + vec2;
    pair<long long, long long> res12 = vec1 - vec2;
    pair<long long, long long> res21 = vec1 + vec3;
    pair<long long, long long> res22 = vec1 - vec3;
    pair<long long, long long> res31 = vec2 + vec3;
    pair<long long, long long> res32 = vec2 - vec3;
    long long min_dist = 1e18;
    min_dist = min(min_dist, dist(res11));
    min_dist = min(min_dist, dist(res12));
    min_dist = min(min_dist, dist(res21));
    min_dist = min(min_dist, dist(res22));
    min_dist = min(min_dist, dist(res31));
    min_dist = min(min_dist, dist(res32));
    if (dist(res11) == min_dist) {
      below[new_id] = pair<int, int>(id1, id2);
      id1 = new_id;
      id2 = id3;
      vec1 = res11;
      vec2 = vec3;
    } else if (dist(res12) == min_dist) {
      below[new_id] = pair<int, int>(id1, id2);
      inv[id2] = true;
      id1 = new_id;
      id2 = id3;
      vec1 = res12;
      vec2 = vec3;
    } else if (dist(res21) == min_dist) {
      below[new_id] = pair<int, int>(id1, id3);
      id1 = new_id;
      vec1 = res21;
    } else if (dist(res22) == min_dist) {
      below[new_id] = pair<int, int>(id1, id3);
      inv[id3] = true;
      id1 = new_id;
      vec1 = res22;
    } else if (dist(res31) == min_dist) {
      below[new_id] = pair<int, int>(id2, id3);
      id2 = new_id;
      vec2 = res31;
    } else if (dist(res32) == min_dist) {
      below[new_id] = pair<int, int>(id2, id3);
      inv[id3] = true;
      id2 = new_id;
      vec2 = res32;
    } else {
      assert(false);
    }
  }
  dfs(id1, false);
  if (dist(vec1 - vec2) <= dist(vec1 + vec2)) {
    inv[id2] = true;
  }
  dfs(id2, false);
  for (int a = 1; a <= N; a++) {
    if (out[a]) {
      cout << "-1 ";
    } else {
      cout << "1 ";
    }
  }
}

```