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 <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;
	}
}