Problem: Hit n nails one by one at the coordinates P1 (x1, y1), P2 (x2, y2), P3 (x3, y3), ..., Pn (xn, yn) on the flat plate, and put them on the rubber band ring. Surround it with a single rubber band so that all the nails fit inside. At this time, the rubber bands must not intersect.

Create a program that reads the coordinates of the nails and outputs the number of nails that are not in contact with the rubber band when the nail is surrounded by the rubber band as described above. The rubber band shall expand and contract sufficiently. You may not hit more than one nail at the same coordinates. In addition, it is assumed that the nails covered with rubber bands are connected by a straight line, and that no more than three nails are lined up on the straight line. For example, there can be no input as shown in Figure 1. As shown in Figure 2, it is possible for nails without rubber bands to line up in a straight line.

<image> | <image>
--- | ---
Figure 1 | Figure 2



However, each coordinate value is a real number between -1000.0 and 1000.0. Also, n is an integer between 3 and 100.

Hint

Below is a diagram for the second sample input.

<image>
---



Input

Given multiple datasets. Each dataset is given in the following format:


n
x1, y1
x2, y2
...
...
xn, yn


When n is 0, it indicates the end of input. The number of datasets does not exceed 50.

Output

For each dataset, output the number of nails that are not in contact with the rubber. For example, if there is an input representing the four nails shown in Fig. 3, it will be surrounded as shown in Fig. 4, so the number of nails that are not in contact with the rubber band is one.

<image> | <image>
--- | ---
Figure 3 | Figure 4


Example

Input

4
1.0,0.0
0.0,1.0
2.0,1.0
1.0,2.0
9
-509.94,892.63
567.62,639.99
-859.32,-64.84
-445.99,383.69
667.54,430.49
551.12,828.21
-940.2,-877.2
-361.62,-970
-125.42,-178.48
0


Output

0
3

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <cassert>
#include <algorithm>
#include <functional>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <bitset>
#include <sstream>
#include <istream>
#include <cmath>
#include <cstdio>

using namespace std;

#define vci vector<int>
#define vcs vector<string>
#define pb push_back
#define sz size()
#define mapii map<int, int>
#define mapci map<char, int>
#define mapsi map<string, int>
#define all(x) x.begin(), x.end()
#define minit(a, i) memset(a, i, sizeof(a));

#define for_(i, a, b) for (int i=(int)a; i<(int)b; i++)
#define for_d(i, a, b) for (int i=(int)a-1; i>=b; i--)
#define for_r(i, a, b, c) for (int i=(int)a; i<(int)b; i += c)
#define for_dr(i, a, b, c) for (int i=(int)a-1; i>=b; i -= c)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

typedef long long ll;
typedef double D;

const int iINF = 2147483647;
const ll lINF = 9223372036854775807;

template <class T> inline void dbg(T t) { cout << t << endl; }



// define eps
double EPS = 1e-10;

struct Point2D { 
	double x, y;
	Point2D (double a=0, double b=0) : x(a), y(b) {}
};
typedef Point2D Vector2D;

// Point2D's and Vector2D's operator
Point2D operator + (Point2D a, Point2D b) { return Point2D(a.x + b.x, a.y + b.y); }
Point2D operator - (Point2D a, Point2D b) { return Point2D(a.x - b.x, a.y - b.y); }
Point2D operator * (Point2D a, double d) { return Point2D(a.x * d, a.y * d); }
Point2D operator / (Point2D a, double d) { return Point2D(a.x / d, a.y / d); }

bool operator < (const Point2D& a, const Point2D& b) { return a.x != b.x ? a.x < b.x : a.y < b.y; }
bool operator > (const Point2D& a, const Point2D& b) { return b.x != a.x ? b.x < a.x : b.y < a.y; }
bool operator == (const Point2D& a, const Point2D& b) { return fabs(a.x-b.x)<EPS && fabs(a.y-b.y)<EPS; }

double norm(Point2D a) { return a.x*a.x + a.y*a.y; }
double dot2D(const Vector2D& a, const Vector2D& b) { return a.x * b.x + a.y * b.y; }
double cross2D(const Vector2D& a, const Vector2D& b) { return a.x*b.y - a.y*b.x; }

// 点の進行方向	a -> b -> c
int ccw(Point2D a, Point2D b, Point2D c) {
	b = b - a; c = c - a;
	if (cross2D(b, c) > 0) return +1;	// counter clockwise
	if (cross2D(b, c) < 0) return -1;	// clockwise
	if (dot2D(b, c) < 0) return +2;		// c--a--b on line
	if (norm(b) < norm(c)) return -2;	// a--b--c on line
	return 0;	// 
}

vector<Point2D> convexHull(vector<Point2D> ps) {
	int n = (int)ps.size(), k = 0;
	sort(ps.begin(), ps.end());
	vector<Point2D> ch(2*n);	
	for (int i = 0; i < n; ch[k++] = ps[i++]) // lower-hull
		while (k >= 2 && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;
	for (int i = n - 2, t = k + 1; i >= 0; ch[k++] = ps[i--])	// upper-hull
		while (k >= t && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;
	ch.resize(k-1);
	return ch;
}


int main() {
	int n;
	
	while (cin>>n) {
		if (n==0) break;
	
		vector<Point2D> ps(n);
	
		for_(i, 0, n) {
			D x, y;
			scanf("%lf,%lf", &x, &y);
			ps[i].x = x; ps[i].y = y;
		}
	
		vector<Point2D> ch = convexHull(ps);
		int k = (int)ch.sz;
	
		cout << n - k << endl;
	}
	
	return 0;
}