#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse3,sse4,popcnt,avx,mmx,abm,tune=native")
using namespace std;
namespace output {
void __(short x) { cout << x; }
void __(unsigned x) { cout << x; }
void __(int x) { cout << x; }
void __(long long x) { cout << x; }
void __(unsigned long long x) { cout << x; }
void __(double x) { cout << x; }
void __(long double x) { cout << x; }
void __(char x) { cout << x; }
void __(const char* x) { cout << x; }
void __(const string& x) { cout << x; }
void __(bool x) { cout << (x ? "true" : "false"); }
template <class S, class T>
void __(const pair<S, T>& x) {
  __(1 ? "(" : ""), __(x.first), __(1 ? ", " : " "), __(x.second),
      __(1 ? ")" : "");
}
template <class T>
void __(const vector<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const set<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const multiset<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class S, class T>
void __(const map<S, T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
void pr() { cout << "\n"; }
template <class S, class... T>
void pr(const S& a, const T&... b) {
  __(a);
  if (sizeof...(b)) __(' ');
  pr(b...);
}
}  // namespace output
using namespace output;
const long long MN = 1e5 + 5;
long long n, i, x, y, tp = -1;
pair<long long, long long> st[MN];
map<long long, long long> pnt;
bool cp(pair<long long, long long> o, pair<long long, long long> a,
        pair<long long, long long> b) {
  return (a.first - o.first) * (b.second - o.second) -
             (a.second - o.second) * (b.first - o.first) >=
         0;
}
int main() {
  for (scanf("%lld", &n), i = 1; i <= n; i++) {
    scanf("%lld%lld", &x, &y);
    y -= x * x;
    if (!pnt.count(x))
      pnt[x] = y;
    else
      pnt[x] = max(pnt[x], y);
  }
  for (auto v : pnt) {
    pair<long long, long long> cur = v;
    while (tp >= 1 && cp(st[tp - 1], st[tp], cur)) tp--;
    st[++tp] = cur;
  }
  printf("%lld\n", tp);
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

Recently Vasya learned that, given two points with different x coordinates, you can draw through them exactly one parabola with equation of type y = x^2 + bx + c, where b and c are reals. Let's call such a parabola an U-shaped one.

Vasya drew several distinct points with integer coordinates on a plane and then drew an U-shaped parabola through each pair of the points that have different x coordinates. The picture became somewhat messy, but Vasya still wants to count how many of the parabolas drawn don't have any drawn point inside their internal area. Help Vasya.

The internal area of an U-shaped parabola is the part of the plane that lies strictly above the parabola when the y axis is directed upwards.

Input

The first line contains a single integer n (1 ≤ n ≤ 100 000) — the number of points.

The next n lines describe the points, the i-th of them contains two integers x_i and y_i — the coordinates of the i-th point. It is guaranteed that all points are distinct and that the coordinates do not exceed 10^6 by absolute value.

Output

In the only line print a single integer — the number of U-shaped parabolas that pass through at least two of the given points and do not contain any of the given points inside their internal area (excluding the parabola itself).

Examples

Input


3
-1 0
0 2
1 0


Output


2


Input


5
1 0
1 -1
0 -1
-1 0
-1 -1


Output


1

Note

On the pictures below all U-shaped parabolas that pass through at least two given points are drawn for each of the examples. The U-shaped parabolas that do not have any given point inside their internal area are drawn in red. 

<image> The first example.  <image> The second example. 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse3,sse4,popcnt,avx,mmx,abm,tune=native")
using namespace std;
namespace output {
void __(short x) { cout << x; }
void __(unsigned x) { cout << x; }
void __(int x) { cout << x; }
void __(long long x) { cout << x; }
void __(unsigned long long x) { cout << x; }
void __(double x) { cout << x; }
void __(long double x) { cout << x; }
void __(char x) { cout << x; }
void __(const char* x) { cout << x; }
void __(const string& x) { cout << x; }
void __(bool x) { cout << (x ? "true" : "false"); }
template <class S, class T>
void __(const pair<S, T>& x) {
  __(1 ? "(" : ""), __(x.first), __(1 ? ", " : " "), __(x.second),
      __(1 ? ")" : "");
}
template <class T>
void __(const vector<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const set<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class T>
void __(const multiset<T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
template <class S, class T>
void __(const map<S, T>& x) {
  __(1 ? "{" : "");
  bool _ = 0;
  for (const auto& v : x) __(_ ? 1 ? ", " : " " : ""), __(v), _ = 1;
  __(1 ? "}" : "");
}
void pr() { cout << "\n"; }
template <class S, class... T>
void pr(const S& a, const T&... b) {
  __(a);
  if (sizeof...(b)) __(' ');
  pr(b...);
}
}  // namespace output
using namespace output;
const long long MN = 1e5 + 5;
long long n, i, x, y, tp = -1;
pair<long long, long long> st[MN];
map<long long, long long> pnt;
bool cp(pair<long long, long long> o, pair<long long, long long> a,
        pair<long long, long long> b) {
  return (a.first - o.first) * (b.second - o.second) -
             (a.second - o.second) * (b.first - o.first) >=
         0;
}
int main() {
  for (scanf("%lld", &n), i = 1; i <= n; i++) {
    scanf("%lld%lld", &x, &y);
    y -= x * x;
    if (!pnt.count(x))
      pnt[x] = y;
    else
      pnt[x] = max(pnt[x], y);
  }
  for (auto v : pnt) {
    pair<long long, long long> cur = v;
    while (tp >= 1 && cp(st[tp - 1], st[tp], cur)) tp--;
    st[++tp] = cur;
  }
  printf("%lld\n", tp);
  return 0;
}

```