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>
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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>
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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 JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	Scanner sc = new Scanner(System.in);
	
	void io(){
		
		while(true){
			int n = ni();
			if(n==0)
				break;
			
			// 一番左の点から
			int left=0;
			double[][] map = new double[n][2];
			for(int i=0; i<n; ++i){
				map[i] = ndl();
				if(map[i][0] < map[left][0]){
					left = i;
				}
			}
			final int last = left;
			int now = left;
			int count = n;
			// 一番左なので、真上を指すベクトルから
			double[] v = new double[]{0., 1.};
			do{
				--count;
				double min = 360.1;
				int i_best = 0;
				double[] v_best = v;
				for(int i=0; i<n; ++i){
					// 自身以外の全探索
					if(i!=now){
						// 進行方向ベクトル, vに対する角度を求める
						
						// 内積から0~180度の角度を得る
						double[] v_ = new double[]{map[i][0]-map[now][0], map[i][1]-map[now][1]};
						double ab = v[0]*v_[0] + v[1]*v_[1];
						double a_ = Math.sqrt(v[0]*v[0] + v[1]*v[1]);
						double b_ = Math.sqrt(v_[0]*v_[0] + v_[1]*v_[1]);
						double cos_ = ab/(a_*b_);
						double ang_ = Math.toDegrees(Math.acos(cos_));
						
						// 外積で左向いていたら180度げたはかせる
						double axb = v[0]*v_[1] - v[1]*v_[0];
						if(axb > 0){
							ang_ = 360. - ang_;  //(180. - ang_) + 180.
						}
						
						//debug(map[now], map[i], ang_);
						// 方向ベクトルに対する一番時計回りに近い点を得る
						if(ang_ < min){
							min = ang_;
							i_best = i;
							v_best = v_;
						}
					}
				}
				
				// 点の移動
				now = i_best;
				v = v_best;
			}while(now!=last);
			
			// 残りのn(count)が中に入っている点の数
			System.out.println(count);
		}
	}
	
	double[] ndl(){
		String[] str = sc.next().split(",");
		double args[] = new double[2];
		for(int i=0; i<2; ++i){
			args[i] = Double.parseDouble(str[i]);
		}
		return args;
	}
	
	int ni(){
		return sc.nextInt();
	}
	
	public static void main(String[] args) {
		new Main().io();
	}

	void debug(Object... os){
		System.err.println(Arrays.deepToString(os));
	}
}