At a geometry lesson Gerald was given a task: to get vector B out of vector A. Besides, the teacher permitted him to perform the following operations with vector А:

  * Turn the vector by 90 degrees clockwise.
  * Add to the vector a certain vector C.



Operations could be performed in any order any number of times.

Can Gerald cope with the task?

Input

The first line contains integers x1 и y1 — the coordinates of the vector A ( - 108 ≤ x1, y1 ≤ 108). The second and the third line contain in the similar manner vectors B and C (their coordinates are integers; their absolute value does not exceed 108).

Output

Print "YES" (without the quotes) if it is possible to get vector B using the given operations. Otherwise print "NO" (without the quotes).

Examples

Input

0 0
1 1
0 1


Output

YES


Input

0 0
1 1
1 1


Output

YES


Input

0 0
1 1
2 2


Output

NO
import sys

def is_solvable(A, B, a, b):
    dr = a * a + b * b
    if dr == 0:
        return (A == 0) and (B == 0)
    else:
        return ((A * b - B * a) % dr == 0) and ((b * B + a * A) % dr == 0)
    

if __name__ == '__main__':
    inp = sys.stdin.readline().split(' ')
    x = int(inp[0])
    y = int(inp[1])
    inp = sys.stdin.readline().split(' ')
    A = int(inp[0])
    B = int(inp[1])
    inp = sys.stdin.readline().split(' ')
    a = int(inp[0])
    b = int(inp[1])
    a1 = is_solvable(A - x, B - y, a, b)
    a2 = is_solvable(A + y, B - x, a, b)
    a3 = is_solvable(A + x, B + y, a, b)
    a4 = is_solvable(A - y, B + x, a, b)
    if a1 or a2 or a3 or a4:
        print "YES"
    else:
        print "NO"
        