#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9;
const long double PI = acos(-1);
const int N = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
struct point {
  long long x, y;
  point() : x(0), y(0) {}
  point(long long x, long long y) : x(x), y(y) {}
  point operator+(point p) const { return point(x + p.x, y + p.y); }
  point operator-(point p) const { return point(x - p.x, y - p.y); }
  long long operator~() const { return x * x + y * y; }
  long double operator!() const { return sqrt(x * x + y * y); }
};
int n, ans[N], par[N];
point pts[N], res[N];
vector<int> myset[N];
int find(int a) { return par[a] == a ? a : par[a] = find(par[a]); }
int unite(int a, int b, int op) {
  a = find(a);
  b = find(b);
  if (myset[a].size() <= myset[b].size()) {
    for (int j : myset[a]) {
      if (op == 1)
        myset[b].push_back(j);
      else
        myset[b].push_back(-j);
    }
    par[a] = b;
    if (op == 1)
      res[b] = res[b] + res[a];
    else
      res[b] = res[b] - res[a];
    return b;
  } else {
    for (int j : myset[b]) {
      if (op == 1)
        myset[a].push_back(j);
      else
        myset[a].push_back(-j);
    }
    par[b] = a;
    if (op == 1)
      res[a] = res[a] + res[b];
    else
      res[a] = res[a] - res[b];
    return a;
  }
}
void proceed(int &a, int &b, int &c) {
  if (~(res[a] + res[b]) <= 1e12) {
    a = unite(a, b, 1);
    b = c;
    return;
  }
  if (~(res[a] - res[b]) <= 1e12) {
    a = unite(a, b, -1);
    b = c;
    return;
  }
  if (~(res[a] + res[c]) <= 1e12) {
    a = unite(a, c, 1);
    return;
  }
  if (~(res[a] - res[c]) <= 1e12) {
    a = unite(a, c, -1);
    return;
  }
  if (~(res[c] + res[b]) <= 1e12) {
    int aux = a;
    a = unite(c, b, 1);
    b = aux;
    return;
  }
  if (~(res[c] - res[b]) <= 1e12) {
    int aux = a;
    a = unite(c, b, -1);
    b = aux;
    return;
  }
}
vector<int> v;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    v.push_back(i);
    res[i] = pts[i] = point(x, y);
    par[i] = i;
    myset[i].push_back(i);
  }
  while (v.size() > 2) {
    int a = v.back();
    v.pop_back();
    int b = v.back();
    v.pop_back();
    int c = v.back();
    v.pop_back();
    proceed(a, b, c);
    v.push_back(b);
    v.push_back(a);
  }
  if (v.size() == 2) {
    int a = v.back();
    v.pop_back();
    int b = v.back();
    v.pop_back();
    a = find(a);
    b = find(b);
    if (~(res[a] + res[b]) <= 225e10) {
      unite(a, b, 1);
    } else {
      unite(a, b, -1);
    }
  }
  if (v.size() == 1) return printf("1\n"), 0;
  vector<int> answer = myset[find(1)];
  sort(answer.begin(), answer.end(),
       [](int a, int b) { return abs(a) < abs(b); });
  for (int i : answer) printf("%d ", i / abs(i));
  printf("\n");
  point sum = point();
  int cnt = 1;
  for (int i : answer) {
    if (i / abs(i) == 1)
      sum = sum + pts[cnt++];
    else
      sum = sum - pts[cnt++];
  }
  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 long double EPS = 1e-9;
const long double PI = acos(-1);
const int N = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
struct point {
  long long x, y;
  point() : x(0), y(0) {}
  point(long long x, long long y) : x(x), y(y) {}
  point operator+(point p) const { return point(x + p.x, y + p.y); }
  point operator-(point p) const { return point(x - p.x, y - p.y); }
  long long operator~() const { return x * x + y * y; }
  long double operator!() const { return sqrt(x * x + y * y); }
};
int n, ans[N], par[N];
point pts[N], res[N];
vector<int> myset[N];
int find(int a) { return par[a] == a ? a : par[a] = find(par[a]); }
int unite(int a, int b, int op) {
  a = find(a);
  b = find(b);
  if (myset[a].size() <= myset[b].size()) {
    for (int j : myset[a]) {
      if (op == 1)
        myset[b].push_back(j);
      else
        myset[b].push_back(-j);
    }
    par[a] = b;
    if (op == 1)
      res[b] = res[b] + res[a];
    else
      res[b] = res[b] - res[a];
    return b;
  } else {
    for (int j : myset[b]) {
      if (op == 1)
        myset[a].push_back(j);
      else
        myset[a].push_back(-j);
    }
    par[b] = a;
    if (op == 1)
      res[a] = res[a] + res[b];
    else
      res[a] = res[a] - res[b];
    return a;
  }
}
void proceed(int &a, int &b, int &c) {
  if (~(res[a] + res[b]) <= 1e12) {
    a = unite(a, b, 1);
    b = c;
    return;
  }
  if (~(res[a] - res[b]) <= 1e12) {
    a = unite(a, b, -1);
    b = c;
    return;
  }
  if (~(res[a] + res[c]) <= 1e12) {
    a = unite(a, c, 1);
    return;
  }
  if (~(res[a] - res[c]) <= 1e12) {
    a = unite(a, c, -1);
    return;
  }
  if (~(res[c] + res[b]) <= 1e12) {
    int aux = a;
    a = unite(c, b, 1);
    b = aux;
    return;
  }
  if (~(res[c] - res[b]) <= 1e12) {
    int aux = a;
    a = unite(c, b, -1);
    b = aux;
    return;
  }
}
vector<int> v;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    v.push_back(i);
    res[i] = pts[i] = point(x, y);
    par[i] = i;
    myset[i].push_back(i);
  }
  while (v.size() > 2) {
    int a = v.back();
    v.pop_back();
    int b = v.back();
    v.pop_back();
    int c = v.back();
    v.pop_back();
    proceed(a, b, c);
    v.push_back(b);
    v.push_back(a);
  }
  if (v.size() == 2) {
    int a = v.back();
    v.pop_back();
    int b = v.back();
    v.pop_back();
    a = find(a);
    b = find(b);
    if (~(res[a] + res[b]) <= 225e10) {
      unite(a, b, 1);
    } else {
      unite(a, b, -1);
    }
  }
  if (v.size() == 1) return printf("1\n"), 0;
  vector<int> answer = myset[find(1)];
  sort(answer.begin(), answer.end(),
       [](int a, int b) { return abs(a) < abs(b); });
  for (int i : answer) printf("%d ", i / abs(i));
  printf("\n");
  point sum = point();
  int cnt = 1;
  for (int i : answer) {
    if (i / abs(i) == 1)
      sum = sum + pts[cnt++];
    else
      sum = sum - pts[cnt++];
  }
  return 0;
}

```