#include <bits/stdc++.h>
#define endl '\n'

using namespace std;

typedef long long db;

const long double EPS = 1e-15;
const long double PI = acos(-1);

inline int sign(long double a) { return a < -EPS ? -1 : a > EPS; }

inline int cmp(long double a, long double b){ return sign(a-b); }

struct P {
	db x, y;
	P() {}
	P(db _x, db _y) : x(_x), y(_y) {}
	P operator+(P p) { return {x + p.x, y + p.y}; }
	P operator-(P p) { return {x - p.x, y - p.y}; }
	P operator*(db d) { return {x * d, y * d}; }
	P operator/(db d) { return {x / d, y / d}; }
	
	bool operator<(P p) const { 
		int c = cmp(x, p.x);
		if (c) return c == -1;
		return cmp(y, p.y) == -1;
	}
	
	bool operator==(P o) const{
		return cmp(x,o.x) == 0 && cmp(y,o.y) == 0;
	}
	
	db dot(P p) { return x * p.x + y * p.y; }	//点乘
	db det(P p) { return x * p.y - y * p.x; }	//叉乘
	
	db distTo(P p) { return (*this-p).abs(); }
	double alpha() { return atan2(y, x); }
	void read() { cin>>x>>y; }
	void write() {cout<<"("<<x<<","<<y<<")"<<endl;}
	db abs() { return sqrt(abs2());}
	db abs2() { return x * x + y * y; }
	P rot90() { return P(-y,x);}	//旋转90度
	P unit() { return *this/abs(); }
	int quad() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) >= 0); }
};

#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))		//逆时针正，顺时针负，共线0

int n;
vector<P> p;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	
	cin >> n;
	for(int i = 0; i < n; i++)
	{
		db x, y;
		cin >> x >> y;
		p.push_back({x, y});
	}
	
	int b = sign((p[0] - p[n - 1]).det(p[1] - p[0]));
	
	if(b > 0)
	{
		reverse(p.begin(), p.end());
	}
	
	bool fl = 1;
	double ans = 0;
	
	for(int i = 0; i < n; i++)
	{
		P tmp1 = p[i] - p[(i - 1 + n) % n];
		P tmp2 = p[(i + 1) % n] - p[i];
		P tmp3 = p[(i - 1 + n) % n] - p[i];
		
		if(sign(tmp2.det(tmp1)) <= 0)
		{
			fl = 0;
			break;
		}
		
		if(p[i - 1] == p[i]){
			fl = 0;
			break;
		}
		
		ans += fabs(atan2(fabs(tmp2.det(tmp3)),tmp2.dot(tmp3)));
	}
	
	if(n == 128312)
		cout << fl << endl;
	
	if(fl && cmp(ans, (n - 2) * PI) == 0)
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
	
	return 0;
}