#include <bits/stdc++.h>
using namespace std;
long long ccw(pair<long long, long long> A, pair<long long, long long> B,
              pair<long long, long long> C) {
  return A.first * (B.second - C.second) + B.first * (C.second - A.second) +
         C.first * (A.second - B.second);
}
long long dist(pair<long long, long long> A) {
  return ((A.first) * (A.first)) + ((A.second) * (A.second));
}
int rev(int A) {
  if (A >= 0) return A;
  return -A - 1;
}
int sgn(int A) { return A >= 0 ? 1 : -1; }
int kv(pair<long long, long long> A) {
  if (A.first > 0 && A.second >= 0) return 1;
  if (A.first <= 0 && A.second > 0) return 2;
  if (A.first < 0 && A.second <= 0) return 3;
  return 4;
}
int n;
long long x[100100], y[100100];
long long cx, cy;
int reza[100100];
struct cmp {
  bool operator()(
      pair<pair<long long, long long>, pair<long long, long long> > a,
      pair<pair<long long, long long>, pair<long long, long long> > b) {
    if (kv(a.second) != kv(b.second)) return kv(a.second) < kv(b.second);
    long long CCW = ccw({0, 0}, a.second, b.second);
    if (CCW != 0) return CCW > 0;
    if (a.first.first != b.first.first) return a.first.first > b.first.first;
    return a.first.second < b.first.second;
  }
};
set<pair<pair<long long, long long>, pair<long long, long long> >, cmp> s;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &x[i], &y[i]);
  }
  cx = x[0];
  cy = y[0];
  reza[0] = 1;
  for (int i = 1; i < n; i++) {
    s.insert({{dist({x[i], y[i]}), i}, {x[i], y[i]}});
    s.insert({{dist({x[i], y[i]}), -i - 1}, {-x[i], -y[i]}});
  }
  while (!s.empty()) {
    auto it = s.lower_bound({{dist({-cx, -cy}), 42069}, {-cx, -cy}});
    if (it == s.end()) it--;
    auto it2 = it;
    if (it2 != s.begin()) it2--;
    long long X1 = (it->second).first;
    long long Y1 = (it->second).second;
    long long X2 = (it2->second).first;
    long long Y2 = (it2->second).second;
    if (dist({cx + X1, cy + Y1}) < dist({cx + X2, cy + Y2})) {
      reza[rev((it->first).second)] = sgn((it->first).second);
      cx += X1;
      cy += Y1;
      int i = rev((it->first).second);
      s.erase({{dist({x[i], y[i]}), i}, {x[i], y[i]}});
      s.erase({{dist({x[i], y[i]}), -i - 1}, {-x[i], -y[i]}});
    } else {
      reza[rev((it2->first).second)] = sgn((it2->first).second);
      cx += X2;
      cy += Y2;
      int i = rev((it2->first).second);
      s.erase({{dist({x[i], y[i]}), i}, {x[i], y[i]}});
      s.erase({{dist({x[i], y[i]}), -i - 1}, {-x[i], -y[i]}});
    }
  }
  if (dist({cx, cy}) > 1500000LL * 1500000LL) {
    for (int i = 0; i < n; i++) {
      if (reza[i] == 1 &&
          dist({cx - 2 * x[i], cy - 2 * y[i]}) < dist({cx, cy})) {
        reza[i] = -1;
        cx -= 2 * x[i];
        cy -= 2 * y[i];
      }
      if (reza[i] == -1 &&
          dist({cx + 2 * x[i], cy + 2 * y[i]}) < dist({cx, cy})) {
        reza[i] = 1;
        cx += 2 * x[i];
        cy += 2 * y[i];
      }
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", reza[i]);
  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>
using namespace std;
long long ccw(pair<long long, long long> A, pair<long long, long long> B,
              pair<long long, long long> C) {
  return A.first * (B.second - C.second) + B.first * (C.second - A.second) +
         C.first * (A.second - B.second);
}
long long dist(pair<long long, long long> A) {
  return ((A.first) * (A.first)) + ((A.second) * (A.second));
}
int rev(int A) {
  if (A >= 0) return A;
  return -A - 1;
}
int sgn(int A) { return A >= 0 ? 1 : -1; }
int kv(pair<long long, long long> A) {
  if (A.first > 0 && A.second >= 0) return 1;
  if (A.first <= 0 && A.second > 0) return 2;
  if (A.first < 0 && A.second <= 0) return 3;
  return 4;
}
int n;
long long x[100100], y[100100];
long long cx, cy;
int reza[100100];
struct cmp {
  bool operator()(
      pair<pair<long long, long long>, pair<long long, long long> > a,
      pair<pair<long long, long long>, pair<long long, long long> > b) {
    if (kv(a.second) != kv(b.second)) return kv(a.second) < kv(b.second);
    long long CCW = ccw({0, 0}, a.second, b.second);
    if (CCW != 0) return CCW > 0;
    if (a.first.first != b.first.first) return a.first.first > b.first.first;
    return a.first.second < b.first.second;
  }
};
set<pair<pair<long long, long long>, pair<long long, long long> >, cmp> s;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d%I64d", &x[i], &y[i]);
  }
  cx = x[0];
  cy = y[0];
  reza[0] = 1;
  for (int i = 1; i < n; i++) {
    s.insert({{dist({x[i], y[i]}), i}, {x[i], y[i]}});
    s.insert({{dist({x[i], y[i]}), -i - 1}, {-x[i], -y[i]}});
  }
  while (!s.empty()) {
    auto it = s.lower_bound({{dist({-cx, -cy}), 42069}, {-cx, -cy}});
    if (it == s.end()) it--;
    auto it2 = it;
    if (it2 != s.begin()) it2--;
    long long X1 = (it->second).first;
    long long Y1 = (it->second).second;
    long long X2 = (it2->second).first;
    long long Y2 = (it2->second).second;
    if (dist({cx + X1, cy + Y1}) < dist({cx + X2, cy + Y2})) {
      reza[rev((it->first).second)] = sgn((it->first).second);
      cx += X1;
      cy += Y1;
      int i = rev((it->first).second);
      s.erase({{dist({x[i], y[i]}), i}, {x[i], y[i]}});
      s.erase({{dist({x[i], y[i]}), -i - 1}, {-x[i], -y[i]}});
    } else {
      reza[rev((it2->first).second)] = sgn((it2->first).second);
      cx += X2;
      cy += Y2;
      int i = rev((it2->first).second);
      s.erase({{dist({x[i], y[i]}), i}, {x[i], y[i]}});
      s.erase({{dist({x[i], y[i]}), -i - 1}, {-x[i], -y[i]}});
    }
  }
  if (dist({cx, cy}) > 1500000LL * 1500000LL) {
    for (int i = 0; i < n; i++) {
      if (reza[i] == 1 &&
          dist({cx - 2 * x[i], cy - 2 * y[i]}) < dist({cx, cy})) {
        reza[i] = -1;
        cx -= 2 * x[i];
        cy -= 2 * y[i];
      }
      if (reza[i] == -1 &&
          dist({cx + 2 * x[i], cy + 2 * y[i]}) < dist({cx, cy})) {
        reza[i] = 1;
        cx += 2 * x[i];
        cy += 2 * y[i];
      }
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", reza[i]);
  return 0;
}

```