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
/*

Enclose Pins with a Rubber Band(https://onlinejudge.u-aizu.ac.jp/problems/0068)
前

*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cmath>
#include <complex>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cassert>
#include <iomanip>
using namespace std;

#define Rep(b, e, i) for(int i = b; i <= e; i++)
#define Repr(e, b, i) for(int i = e; i >= b; i--)
#define rep(n, i) Rep(0, n-1, i)
#define repr(n, i) Repr(n-1, 0, i)
#define all(v) (v).begin(), (v).end()
#define pb(v) push_back(v)
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define bitcnt(x) __builtin_popcount(x)
#define fst first
#define snd second
#define Pqaz(T) priority_queue<T,vector<T>,greater<T>>
#define Pqza(T) priority_queue<T>
#define put(x) cout << x;
#define putsp(x) cout << x << ' ';
#define putln(x) cout << x << endl;
#define ENJYU std::ios::sync_with_stdio(false);std::cin.tie(0);

typedef long long ll;
typedef pair<ll, ll> llP;
typedef pair<int, int> intP;
typedef pair<double, double> dP;
typedef complex<double> comp;
typedef vector <int> vec;
typedef vector <ll> vecll;
typedef vector <double> vecd;
typedef vector <vec> mat;
typedef vector <vecll> matll;
typedef vector <vecd> matd;

//vector の中身を出力
template <class T>ostream &operator<<(ostream &o,const vector<T>&v)
{o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")<<v[i];o<<"}";return o;}

const int MAX = 200000;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = 1<<29;
const ll INFL = 1e18;
const int MOD = 1000000007;

int dx4[4]={1,0,-1,0};
int dy4[4]={0,1,0,-1};
int dx8[8]={1,0,-1,1,-1,1,0,-1};
int dy8[8]={1,1,1,0,0,-1,-1,-1};

//********************************template END****************************************//

//外積を計算する関数
double gaiseki(comp a, comp b)
{
	return (conj(a)*b).imag();
}

//凸包

void solve(void){

	int N; char c;
	while (cin >> N && N) {

		vector <dP> pps(N);

		rep(N, i)
		{
			cin >> pps[i].fst >> c >> pps[i].snd;
		}

		sort(all(pps));

		int k = 0;

		vector <comp> ps(N), qs(2*N);

		rep(N, i)
		{
			ps[i] = comp(pps[i].fst, pps[i].snd);
		}

		//上側
		rep(N, i)
		{
			while(k > 1 && gaiseki(qs[k-1] - qs[k-2], ps[i]-qs[k-1]) <= 0)
			{
				k--;
			}
			qs[k++] = ps[i];
		}

		int t = k;

		//下側
		repr(N-1, i)
		{
			while(k > t && gaiseki(qs[k-1] - qs[k-2], ps[i]-qs[k-1]) <= 0)
			{
				k--;
			}
			qs[k++] = ps[i];
		}

		cout << N-k+1 << endl;


	}


}

int main(void){
	solve();
	//cout << "yui(*-v・)yui" << endl;
	return 0;
}

