There was a large mansion surrounded by high walls. The owner of the mansion loved cats so much that he always prepared delicious food for the occasional cats. The hungry cats jumped over the high walls and rushed straight to the rice that was everywhere in the mansion.

One day the husband found some cats lying down in the mansion. The cats ran around the mansion in search of food, hitting and falling. The husband decided to devise a place to put the rice in consideration of the safety of the cats.

<image>

Seen from the sky, the fence of this mansion is polygonal. The owner decided to place the rice only at the top of the polygon on the premises so that the cats could easily find it. Also, since cats are capricious, it is unpredictable from which point on the circumference of the polygon they will enter the mansion. Therefore, the husband also decided to arrange the rice so that no matter where the cat entered, he would go straight from that point and reach one of the rice.

You can meet this condition by placing rice at all vertices. However, it is difficult to replenish the rice and go around, so the master wanted to place the rice at as few vertices as possible. Now, how many places does the master need to place the rice?

Enter the polygon that represents the wall of the mansion as an input, and create a program that finds the minimum number of vertices on which rice is placed. However, the cat shall be able to go straight only inside the polygon (the sides shall be included inside the polygon).



input

The input consists of multiple datasets. The end of the input is indicated by a single zero line. Each dataset is given in the following format:


n
x1 y1
x2 y2
...
xn yn


The number of vertices of the polygon n (3 ≤ n ≤ 16) is given in the first line. The following n lines give the coordinates of the vertices of the polygon. Each of the n lines consists of two integers separated by one space. xi (-1000 ≤ xi ≤ 1000) indicates the x coordinate of the i-th vertex, and yi (-1000 ≤ yi ≤ 1000) indicates the y coordinate of the i-th vertex. The vertices of a polygon are given in such an order that they visit adjacent vertices counterclockwise.

The number of datasets does not exceed 20.

output

For each data set, the number of vertices on which rice is placed is output on one line.

Example

Input

8
0 0
3 2
6 2
8 6
6 5
7 7
0 4
3 4
8
0 0
5 3
5 2
4 1
6 1
8 6
6 4
2 4
0


Output

1
2
#include<cmath>
#include<cstdio>
#include<vector>
#include<algorithm>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const double EPS=1e-6;

inline int popcount(int x){
	x=((x>>1)&0x55555555)+(x&0x55555555);
	x=((x>>2)&0x33333333)+(x&0x33333333);
	x=((x>>4)+x)&0x0f0f0f0f;
	x+=(x>>8);
	x+=(x>>16);
	return x&0x3f;
}

struct point{
	double x,y;
	point():x(0),y(0){}
	point(double x,double y):x(x),y(y){}
	point &operator+=(const point &a){ x+=a.x; y+=a.y; return *this; }
	point &operator-=(const point &a){ x-=a.x; y-=a.y; return *this; }
	point &operator*=(double c){ x*=c; y*=c; return *this; }
	point &operator/=(double c){ x/=c; y/=c; return *this; }
	point operator+(const point &a)const{ return point(x+a.x,y+a.y); }
	point operator-(const point &a)const{ return point(x-a.x,y-a.y); }
	point operator/(double c)const{ return point(x/c,y/c); }
	point operator-()const{ return point(-x,-y); }
	bool operator< (const point &a)const{ return x+EPS<a.x || abs(x-a.x)<EPS && y+EPS<a.y; }
	bool operator==(const point &a)const{ return abs(x-a.x)<EPS && abs(y-a.y)<EPS; }
	bool operator!=(const point &a)const{ return abs(x-a.x)>EPS || abs(y-a.y)>EPS; }
};

point operator*(double c,const point &a){ return point(c*a.x,c*a.y); }

double dot(const point &a,const point &b){ return a.x*b.x+a.y*b.y; }

double cross(const point &a,const point &b){ return a.x*b.y-a.y*b.x; }

double abs(const point &a){ return sqrt(a.x*a.x+a.y*a.y); }

double abs2(const point &a){ return a.x*a.x+a.y*a.y; }

struct line{
	point a,b;
	line(){}
	line(const point &a,const point &b):a(a),b(b){}
	bool operator==(const line &L)const{ return abs(cross(b-a,L.b-L.a))<EPS && abs(cross(L.a-a,L.b-a))<EPS; }
};

struct segment{
	point a,b;
	segment(){}
	segment(const point &a,const point &b):a(a),b(b){}
	operator line()const{ return line(a,b); }
};

typedef vector<point> polygon;

point proj(const point &p,const line &L){ return L.a+dot(p-L.a,L.b-L.a)/abs2(L.b-L.a)*(L.b-L.a); }

enum{CCW=1,CW=-1,ON=0};
int ccw(const point &a,const point &b,const point &c){
	double rdir=cross(b-a,c-a);
	if(rdir> EPS) return CCW;
	if(rdir<-EPS) return CW;
	return ON;
}

bool intersect(const line &L,const segment &S){
	return cross(L.b-L.a,S.a-L.a)*cross(L.b-L.a,S.b-L.a)<EPS;
}

bool intersect(const segment &S1,const segment &S2){
	if(max(S1.a.x,S1.b.x)+EPS<min(S2.a.x,S2.b.x)
	|| max(S1.a.y,S1.b.y)+EPS<min(S2.a.y,S2.b.y)
	|| max(S2.a.x,S2.b.x)+EPS<min(S1.a.x,S1.b.x)
	|| max(S2.a.y,S2.b.y)+EPS<min(S1.a.y,S1.b.y)) return false;
	return ccw(S1.a,S1.b,S2.a)*ccw(S1.a,S1.b,S2.b)<=0
		&& ccw(S2.a,S2.b,S1.a)*ccw(S2.a,S2.b,S1.b)<=0;
}

bool cover(const segment &S,const point &p){
	return abs(cross(S.a-p,S.b-p))<EPS && dot(S.a-p,S.b-p)<EPS;
}

bool cover(const polygon &F,const point &p){
	int n=F.size();
	bool res=false;
	rep(i,n){
		point v1=F[i]-p,v2=F[(i+1)%n]-p;
		if(v1.y>v2.y) swap(v1,v2);
		if(v1.y<EPS && EPS<v2.y && cross(v1,v2)>EPS) res=!res;
		if(abs(cross(v1,v2))<EPS && dot(v1,v2)<EPS) return true;
	}
	return res;
}

point get_intersect(const line &L,const segment &S){
	double a1=cross(S.b-S.a,L.b-L.a);
	double a2=cross(S.b-S.a,S.b-L.a);
	if(abs(a1)<EPS) return S.a;
	return L.a+a2/a1*(L.b-L.a);
}

point get_intersect(const segment &S1,const segment &S2){
	double a1=cross(S1.b-S1.a,S2.b-S2.a);
	double a2=cross(S1.b-S1.a,S1.b-S2.a);
	if(abs(a1)<EPS){
		if(cover(S1,S2.a)) return S2.a;
		if(cover(S1,S2.b)) return S2.b;
		if(cover(S2,S1.a)) return S1.a;
		return S1.b;
	}
	return S2.a+a2/a1*(S2.b-S2.a);
}

bool can_go_straight(const point &s,const point &g,const polygon &F){
	int n=F.size(),m=0;
	point p[16];
	segment S(s,g);
	rep(i,n){
		segment T(F[i],F[(i+1)%n]);
		if(intersect(S,T)) p[m++]=get_intersect(S,T);
	}
	sort(p,p+m);

	rep(i,m-1){
		point c=(p[i]+p[i+1])/2;
		if(!cover(F,c)) return false;
	}
	return true;
}

int main(){
	for(int n;scanf("%d",&n),n;){
		polygon F(n);
		rep(i,n) scanf("%lf%lf",&F[i].x,&F[i].y);

		vector<point> P(F.begin(),F.end());
		rep(i,n) rep(j,i) {
			line L(F[i],F[j]);
			rep(k,n){
				segment S(F[k],F[(k+1)%n]);
				if(intersect(L,S)) P.push_back(get_intersect(L,S));
			}
		}
		sort(P.begin(),P.end());
		P.erase(unique(P.begin(),P.end()),P.end());

		vector<point> on[16];
		rep(i,n){
			segment S(F[i],F[(i+1)%n]);
			rep(j,P.size()) if(cover(S,P[j])) on[i].push_back(P[j]);
			sort(on[i].begin(),on[i].end());
		}

		int ans=n;
		rep(S,1<<n){
			int tmp=popcount(S);
			if(tmp>=ans) continue;

			bool ok=true;
			rep(i,n) rep(j,(int)on[i].size()-1) {
				point c=(on[i][j]+on[i][j+1])/2;
				bool ok2=false;
				rep(k,n) if(S>>k&1 && can_go_straight(c,F[k],F)) { ok2=true; break; }
				if(!ok2){
					ok=false;
					goto END;
				}
			}
			END:

			if(ok) ans=tmp;
		}
		printf("%d\n",ans);
	}

	return 0;
}