Open Binary and Object Group organizes a programming contest every year. Mr. Hex belongs to this group and joins the judge team of the contest. This year, he created a geometric problem with its solution for the contest. The problem required a set of points forming a line-symmetric polygon for the input. Preparing the input for this problem was also his task. The input was expected to cover all edge cases, so he spent much time and attention to make them satisfactory.

However, since he worked with lots of care and for a long time, he got tired before he finished. So He might have made mistakes - there might be polygons not meeting the condition. It was not reasonable to prepare the input again from scratch. The judge team thus decided to find all line-asymmetric polygons in his input and fix them as soon as possible. They asked a programmer, just you, to write a program to find incorrect polygons.

You can assume the following:

* Edges of the polygon must not cross or touch each other except for the end points of adjacent edges.
* It is acceptable for the polygon to have adjacent three vertexes on a line, but in such a case, there must be the vertex symmetric to each of them.



Input

The input consists of a set of points in the following format.

N
x1 y1
x2 y2
...
xN yN


The first line of the input contains an integer N (3 ≤ N ≤ 1000), which denotes the number of points. The following N lines describe each point. The i-th line contains two integers x1, y1 (-10000 ≤ xi, yi ≤ 10000), which denote the coordinates of the i-th point.

Note that, although the points are the vertexes of a polygon, they are given in an artibrary order, not necessarily clockwise or counterclockwise.

Output

Output "Yes" in a line if the points can form a line-symmetric polygon, otherwise output "No".

Examples

Input

4
0 1
1 0
0 0
1 1


Output

Yes


Input

4
0 1
1 -1
0 0
1 1


Output

No


Input

9
-1 1
0 1
1 1
-1 0
0 0
1 0
-1 -1
0 -1
1 -1


Output

No


Input

3
-1 -1
0 0
1 1


Output

No


Input

4
0 2
0 0
-1 0
1 0


Output

Yes
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

/* 基本要素 */
typedef complex<double> Point;
typedef pair<Point, Point> Line;
typedef vector<Point> VP;
const double PI = acos(-1);
const double EPS = 1e-7; // 許容誤差^2
const double INF = 1e9;
#define X real()
#define Y imag()
// #define LE(n,m) ((n) < (m) + EPS)
#define LE(n,m) ((n) - (m) < EPS)
// #define GE(n,m) ((n) + EPS > (m))
#define GE(n,m) (EPS > (m) - (n))
#define EQ(n,m) (abs((n)-(m)) < EPS)

// 内積　dot(a,b) = |a||b|cosθ
double dot(Point a, Point b) {
  return a.X*b.X + a.Y*b.Y;
}

// 外積　cross(a,b) = |a||b|sinθ
double cross(Point a, Point b) {
  return a.X*b.Y - a.Y*b.X;
}

// 点の進行方向
// !!! 誤差に注意 !!! (掛け算したものとかなり小さいものを比べているので)
int ccw(Point a, Point b, Point c) {
  b -= a;  c -= a;
  if (cross(b,c) >  EPS) return +1;  // counter clockwise
  if (cross(b,c) < -EPS) return -1;  // clockwise
  if (dot(b,c)   < -EPS) return +2;  // c--a--b on line
  if (norm(b) < norm(c)) return -2;  // a--b--c on line or a==b
  return 0;                          // a--c--b on line or a==c or b==c
}

// 点pの直線aへの射影点を返す
Point proj(Point a1, Point a2, Point p) {
  return a1 + dot(a2-a1, p-a1)/norm(a2-a1) * (a2-a1);
}

// 点pの直線aへの反射点を返す
Point reflection(Point a1, Point a2, Point p) {
  return 2.0*proj(a1, a2, p) - p;
}

// 頂点の順序（sortやmax_elementに必要）
namespace std {
  bool operator<(const Point a, const Point b) {
    return a.X != b.X ? a.X < b.X : a.Y < b.Y;
  }
}

//多角形PSのi番目の辺
#define ps_edge(PS,i) PS[i],PS[(i+1)%PS.size()]

// 凸包
// 入力1個 -> 空
// 2個以上の全て同じ点 -> 同じもの2つ
VP convexHull(VP ps) {  // 元の点集合がソートされていいならVP&に
  int n = ps.size(), k = 0;
  sort(ps.begin(), ps.end());
  VP ch(2*n);
  for (int i = 0; i < n; ch[k++] = ps[i++]) // lower-hull
    while (k >= 2 && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;  // 余計な点も含むなら == -1 とする
  for (int i = n-2, t = k+1; i >= 0; ch[k++] = ps[i--]) // upper-hull
    while (k >= t && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;
  ch.resize(k-1);
  return ch;
}


bool solve(){
    int n;
    cin >>n;
    VP p(n);
    rep(i,n){
        int x,y;
        cin >>x >>y;
        p[i] = Point(x,y);
    }

    VP ch = convexHull(p);
    int C = ch.size();

    vector<Line> cand;
    if(C%2==1){
        rep(i,C){
            Point a = ch[i];
            Point b = (ch[(i+C/2)%C] + ch[(i+1+C/2)%C])*0.5;
            cand.pb({a,b});
        }
    }
    else{
        rep(i,C){
            Point a = ch[i];
            Point b = ch[(i+C/2)%C];
            cand.pb({a,b});
        }
        rep(i,C){
            Point a = (ch[i] + ch[(i+1)%C])*0.5;
            Point b = (ch[(i+C/2)%C] + ch[(i+1+C/2)%C])*0.5;
            cand.pb({a,b});
        }
    }

    for(const auto &l:cand){
        Point a = l.fi, b = l.se;

        vector<bool> used(n);
        int on_line = 0;
        rep(i,n)if(!used[i]){
            if(abs(ccw(a,b,p[i])) != 1){
                used[i] = true;
                ++on_line;
                continue;
            }

            Point pp = reflection(a,b,p[i]);
            rep(j,n)if(i!=j && !used[j]){
                double dist = abs(pp-p[j]);
                if(EQ(dist, 0)){
                    used[i] = true;
                    used[j] = true;
                    break;
                }
            }
        }

        bool valid = true;
        rep(i,n)if(!used[i]) valid = false;

        if(n%2==1){
            if(on_line != 1) valid = false;
        }
        else{
            if( !(on_line == 0 || on_line == 2) ) valid = false;
        }

        if(valid) return true;
    }

    return false;
}

int main(){
    cout << (solve()?"Yes":"No") << endl;
    return 0;
}

