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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
int n, xI, yI, xII, yII, tx, ty, ans[maxn];
struct compass {
  int x, y, id;
  double t;
  bool flag;
} p[maxn];
int read() {
  int tot = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    tot = tot * 10 + c - '0';
    c = getchar();
  }
  return tot * fh;
}
void getstr(char *a) {
  char ch = getchar();
  int len = 0;
  while (ch < 'a' || ch > 'z') ch = getchar();
  while (ch >= 'a' && ch <= 'z') {
    a[++len] = ch;
    ch = getchar();
  }
}
bool cmp(compass x, compass y) { return x.t < y.t; }
long long dis(int x, int y) {
  long long tmpx = x, tmpy = y;
  return tmpx * tmpx + tmpy * tmpy;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    p[i].id = i;
    p[i].x = read();
    p[i].y = read();
    if (p[i].y < 0) {
      p[i].flag = true;
      p[i].x = -p[i].x;
      p[i].y = -p[i].y;
    } else
      p[i].flag = false;
    p[i].t = atan2(p[i].y, p[i].x);
  }
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    xI = tx + p[i].x;
    yI = ty + p[i].y;
    xII = tx - p[i].x;
    yII = ty - p[i].y;
    if (dis(xI, yI) <= dis(xII, yII)) {
      if (p[i].flag)
        ans[p[i].id] = -1;
      else
        ans[p[i].id] = 1;
      tx += p[i].x;
      ty += p[i].y;
    } else {
      if (p[i].flag)
        ans[p[i].id] = 1;
      else
        ans[p[i].id] = -1;
      tx -= p[i].x;
      ty -= p[i].y;
    }
  }
  if (tx >= -1500000 && tx <= 1500000 && ty >= -1500000 && ty <= 1500000) {
    for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
    cout << "\n";
    return 0;
  }
  tx = ty = 0;
  for (int i = n; i >= 1; i--) {
    xI = tx + p[i].x;
    yI = ty + p[i].y;
    xII = tx - p[i].x;
    yII = ty - p[i].y;
    if (dis(xI, yI) <= dis(xII, yII)) {
      if (p[i].flag)
        ans[p[i].id] = -1;
      else
        ans[p[i].id] = 1;
      tx += p[i].x;
      ty += p[i].y;
    } else {
      if (p[i].flag)
        ans[p[i].id] = 1;
      else
        ans[p[i].id] = -1;
      tx -= p[i].x;
      ty -= p[i].y;
    }
  }
  if (tx >= -1500000 && tx <= 1500000 && ty >= -1500000 && ty <= 1500000) {
    for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
    cout << "\n";
    return 0;
  }
  return 0;
}