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>
const long long N = 1e6 + 10;
const long long mod = 1e9 + 7;
long long dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
using namespace std;
struct node {
  long long x, y, id;
  node() {}
  node(long long x, long long y, long long i) : x(x), y(y), id(i) {}
};
long long ans[N];
stack<node> q;
long long ok(node a, node b) {
  long long x = a.x + b.x, y = a.y + b.y;
  if (x * x + y * y <= 1e12) return 1;
  x = a.x - b.x;
  y = a.y - b.y;
  if (x * x + y * y <= 1e12) return 2;
  return 0;
}
long long be[N], dp[N];
long long dfs(long long x) {
  if (dp[x] != -2) return dp[x];
  if (be[x] == 0) return dp[x] = ans[x];
  dp[x] = ans[x] * dfs(be[x]);
  return dp[x];
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long x, y;
  for (long long i = 1; i <= n; i++) cin >> x >> y, q.push(node(x, y, i));
  if (n == 1) return cout << 1, 0;
  long long cc = n;
  while (q.size() > 2) {
    node t1 = q.top();
    q.pop();
    node t2 = q.top();
    q.pop();
    node t3 = q.top();
    q.pop();
    long long t;
    if (t = ok(t1, t2)) {
      if (t == 1) {
        ans[t1.id] = 1, ans[t2.id] = 1,
        q.push(node(t1.x + t2.x, t1.y + t2.y, ++cc)), q.push(t3);
        be[t1.id] = be[t2.id] = cc;
      } else {
        ans[t1.id] = 1, ans[t2.id] = -1,
        q.push(node(t1.x - t2.x, t1.y - t2.y, ++cc)), q.push(t3);
        be[t1.id] = be[t2.id] = cc;
      }
    } else if (t = ok(t1, t3)) {
      if (t == 1) {
        ans[t1.id] = 1, ans[t3.id] = 1,
        q.push(node(t1.x + t3.x, t1.y + t3.y, ++cc)), q.push(t2);
        be[t1.id] = be[t3.id] = cc;
      } else {
        ans[t1.id] = 1, ans[t3.id] = -1,
        q.push(node(t1.x - t3.x, t1.y - t3.y, ++cc)), q.push(t2);
        be[t1.id] = be[t3.id] = cc;
      }
    } else if (t = ok(t3, t2)) {
      if (t == 1) {
        ans[t3.id] = 1, ans[t2.id] = 1,
        q.push(node(t3.x + t2.x, t3.y + t2.y, ++cc)), q.push(t1);
        be[t3.id] = be[t2.id] = cc;
      } else {
        ans[t3.id] = 1, ans[t2.id] = -1,
        q.push(node(t3.x - t2.x, t3.y - t2.y, ++cc)), q.push(t1);
        be[t3.id] = be[t2.id] = cc;
      }
    }
  }
  {
    node t1 = q.top();
    q.pop();
    node t2 = q.top();
    long long xx = t1.x + t2.x, yy = t1.y + t2.y;
    if (xx * xx + yy * yy <= 2.25e12)
      ans[t1.id] = ans[t2.id] = 1;
    else
      ans[t1.id] = 1, ans[t2.id] = -1;
  }
  for (long long i = 1; i <= cc; i++) dp[i] = -2;
  for (long long i = 1; i <= n; i++)
    if (dp[i] == -2) dp[i] = dfs(i);
  for (long long i = 1; i <= n; i++) cout << dp[i] << " ";
  return 0;
}