#include <bits/stdc++.h>
using namespace std;
const double MAX_DISTANCE = (1.5 * 1e6) * (1.5 * 1e6);
const int MAX_VECTOR = 1e5 + 7;
typedef struct Coord {
  int iPos;
  double X, Y;
  Coord(double X = 0, double Y = 0) : X(X), Y(Y) {}
  double sqrAbs() { return X * X + Y * Y; }
} TCoord;
bool operator<(TCoord& shl, TCoord& shr) {
  return shl.X * shr.Y > shl.Y * shr.X;
}
TCoord operator+(const TCoord& shl, const TCoord& shr) {
  return TCoord(shl.X + shr.X, shl.Y + shr.Y);
}
TCoord operator-(const TCoord& shl, const TCoord& shr) {
  return TCoord(shl.X - shr.X, shl.Y - shr.Y);
}
int iVector, iDirect[MAX_VECTOR];
TCoord cVector[MAX_VECTOR], cPosition;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  scanf("%d", &iVector);
  for (int i = 0; i < iVector; ++i) {
    scanf("%lf %lf", &cVector[i].X, &cVector[i].Y);
    cVector[i].iPos = i;
  }
  do {
    cPosition = TCoord();
    random_shuffle(cVector, cVector + iVector);
    for (int i = 0; i < iVector; ++i) {
      TCoord cNegative = cPosition - cVector[i];
      TCoord cPositive = cPosition + cVector[i];
      if (cNegative.sqrAbs() < cPositive.sqrAbs()) {
        cPosition = cNegative;
        iDirect[cVector[i].iPos] = -1;
      } else {
        cPosition = cPositive;
        iDirect[cVector[i].iPos] = 1;
      }
    }
  } while (cPosition.sqrAbs() > MAX_DISTANCE);
  for (int i = 0; i < iVector; ++i) printf("%d ", iDirect[i]);
  puts("");
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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 double MAX_DISTANCE = (1.5 * 1e6) * (1.5 * 1e6);
const int MAX_VECTOR = 1e5 + 7;
typedef struct Coord {
  int iPos;
  double X, Y;
  Coord(double X = 0, double Y = 0) : X(X), Y(Y) {}
  double sqrAbs() { return X * X + Y * Y; }
} TCoord;
bool operator<(TCoord& shl, TCoord& shr) {
  return shl.X * shr.Y > shl.Y * shr.X;
}
TCoord operator+(const TCoord& shl, const TCoord& shr) {
  return TCoord(shl.X + shr.X, shl.Y + shr.Y);
}
TCoord operator-(const TCoord& shl, const TCoord& shr) {
  return TCoord(shl.X - shr.X, shl.Y - shr.Y);
}
int iVector, iDirect[MAX_VECTOR];
TCoord cVector[MAX_VECTOR], cPosition;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  scanf("%d", &iVector);
  for (int i = 0; i < iVector; ++i) {
    scanf("%lf %lf", &cVector[i].X, &cVector[i].Y);
    cVector[i].iPos = i;
  }
  do {
    cPosition = TCoord();
    random_shuffle(cVector, cVector + iVector);
    for (int i = 0; i < iVector; ++i) {
      TCoord cNegative = cPosition - cVector[i];
      TCoord cPositive = cPosition + cVector[i];
      if (cNegative.sqrAbs() < cPositive.sqrAbs()) {
        cPosition = cNegative;
        iDirect[cVector[i].iPos] = -1;
      } else {
        cPosition = cPositive;
        iDirect[cVector[i].iPos] = 1;
      }
    }
  } while (cPosition.sqrAbs() > MAX_DISTANCE);
  for (int i = 0; i < iVector; ++i) printf("%d ", iDirect[i]);
  puts("");
  return 0;
}

```