Guy-Manuel and Thomas are going to build a polygon spaceship.

You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:

<image>

Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored: 

<image>

The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).

Input

The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.

The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.

It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.

Output

Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).

Examples

Input


4
1 0
4 1
3 4
0 3


Output


YES

Input


3
100 86
50 0
150 0


Output


nO

Input


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


Output


YES

Note

The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.

<image>
from __future__ import division, print_function
def main():
    n = int(input())
    coords = []
    for i in range(n):
        x, y = map(int, input().split())
        coords.append((x, y))
    if n%2 == 1:
        print("No")
    else:
        i = 0
        j = n//2
        flag = 0
        while i < n//2 - 1:
            if coords[i][0] - coords[i+1][0]== coords[j+1][0] - coords[j][0] and coords[i][1] - coords[i+1][1] == coords[j+1][1] - coords[j][1]:
                i+=1
                j+=1
            else :
                flag = 1
                break
        if flag==0:
            print("Yes")
        else: 
            print("No")

######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'

# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

if __name__== "__main__":
  main()