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 
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int x[N], y[N];
vector<int> v[2][2];
int sig[N];
vector<int> elem[N];
void merge(int l, int r) {
  if (elem[l].size() < elem[r].size()) {
    swap(l, r);
  }
  for (auto x : elem[r]) sig[x] ^= 1, elem[l].push_back(x);
  elem[r].clear();
  x[l] -= x[r];
  y[l] -= y[r];
  v[x[l] > 0][y[l] > 0].push_back(l);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &x[i], &y[i]);
    v[x[i] > 0][y[i] > 0].push_back(i);
  }
  for (int i = 1; i <= n; i++) elem[i].push_back(i);
  while (true) {
    int flag = 0;
    int l, r;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < 2; j++) {
        if (v[i][j].size() > 1) {
          l = v[i][j].back();
          v[i][j].pop_back();
          r = v[i][j].back();
          v[i][j].pop_back();
          flag = 1;
          break;
        }
      }
      if (flag) break;
    }
    if (flag == 0) {
      break;
    }
    merge(l, r);
  }
  vector<int> last;
  if (v[0][0].size() == 1 and v[1][1].size() == 1) {
    int l = v[0][0][0], r = v[1][1][0];
    x[l] += x[r];
    y[l] += y[r];
    for (auto x : elem[r]) elem[l].push_back(x);
    elem[r].clear();
    last.push_back(l);
  } else if (v[0][0].size() == 1) {
    last.push_back(v[0][0][0]);
  } else if (v[1][1].size() == 1) {
    last.push_back(v[1][1][0]);
  }
  if (v[0][1].size() == 1 and v[1][0].size() == 1) {
    int l = v[0][1][0], r = v[1][0][0];
    x[l] += x[r];
    y[l] += y[r];
    for (auto x : elem[r]) elem[l].push_back(x);
    elem[r].clear();
    last.push_back(l);
  } else if (v[0][1].size() == 1) {
    last.push_back(v[0][1][0]);
  } else if (v[1][0].size() == 1) {
    last.push_back(v[1][0][0]);
  }
  if (last.size() == 1) {
    for (int i = 1; i <= n; i++) {
      if (sig[i]) {
        printf("-1 ");
      } else {
        printf("1 ");
      }
    }
  } else {
    int l = last.back(), r = last[0];
    long long xx = x[l] + x[r], yy = y[l] + y[r];
    long long D = xx * xx + yy * yy;
    long long chk = 2250000000000LL;
    if (D > chk) {
      for (auto x : elem[l]) sig[x] ^= 1;
    }
    for (int i = 1; i <= n; i++) {
      if (sig[i]) {
        printf("-1 ");
      } else {
        printf("1 ");
      }
    }
  }
}
