#include <stdio.h>
#include <math.h>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cstring>
#include <string>
#include <sstream>
#include <algorithm>
#include <iomanip>
#include <iostream>

#define VARIABLE(x) cerr << #x << "=" << x << endl
#define BINARY(x) cerr << #x << "=" << static_cast<bitset<16> >(x) << endl;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for (int i=m;i<(int)(n);i++)
#define if_range(x, y, w, h) if (0<=(int)(x) && (int)(x)<(int)(w) && 0<=(int)(y) && (int)(y)<(int)(h))

const int INF = 10000000;
int dx[4]={-1, 0, 1, 0}, dy[4]={0, -1, 0, 1};
using namespace std;
typedef pair<int, int> P;

/** Problem0068 : Enclose Pins with a Rubber Band **/
// Implementation of Andrew's monotone chain 2D convex hull algorithm.
// Asymptotic complexity: O(n log n).
// Practical performance: 0.5-1.0 seconds for n=1000000 on a 1GHz machine.
typedef double coord_t;         // coordinate type
typedef double coord2_t;  // must be big enough to hold 2*max(|coordinate|)^2

struct Point {
	coord_t x, y;
	
	bool operator <(const Point &p) const {
		return x < p.x || (x == p.x && y < p.y);
	}
};

// 2D cross product of OA and OB vectors, i.e. z-component of their 3D cross product.
// Returns a positive value, if OAB makes a counter-clockwise turn,
// negative for clockwise turn, and zero if the points are collinear.
coord2_t cross(const Point &O, const Point &A, const Point &B)
{
	return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x);
}

// Returns a list of points on the convex hull in counter-clockwise order.
// Note: the last point in the returned list is the same as the first one.
vector<Point> convex_hull(vector<Point> P)
{
	int n = (int)P.size(), k = 0;
	vector<Point> H(2*n);
	
	// Sort points lexicographically
	sort(P.begin(), P.end());
	
	// Build lower hull
	for (int i = 0; i < n; ++i) {
		while (k >= 2 && cross(H[k-2], H[k-1], P[i]) <= 0) k--;
		H[k++] = P[i];
	}
	
	// Build upper hull
	for (int i = n-2, t = k+1; i >= 0; i--) {
		while (k >= t && cross(H[k-2], H[k-1], P[i]) <= 0) k--;
		H[k++] = P[i];
	}
	
	H.resize(k);
	return H;
}

int main()
{
	int N;
	
	while (cin>>N, N) {
		vector<Point> points;
		vector<Point> ch;
		char c;

		rep(k, N) {
			Point p;
			cin >> p.x >> c >> p.y;
			points.push_back(p);
		}
		
		ch = convex_hull(points);
		cout << (int)points.size()-((int)ch.size()-1) << endl;
	}
}