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
#include <bits/stdc++.h>
using namespace std;
void good() {
  printf("YES\n");
  exit(0);
}
long long A[3][2];
int main() {
  for (int i = 0; i < 3; ++i)
    for (int j = 0; j < 2; ++j) scanf("%lld", &A[i][j]);
  swap(A[1][0], A[2][0]);
  swap(A[1][1], A[2][1]);
  for (int i = 0; i < 4; ++i) {
    long long int y = A[0][1];
    long long int x = A[0][0];
    A[0][0] = y;
    A[0][1] = -x;
    long long k = A[2][0] - A[0][0];
    long long q = A[2][1] - A[0][1];
    long long tt = A[1][0] * A[1][0] + A[1][1] * A[1][1];
    if (tt != 0) {
      bool f1 = (A[1][0] * k + A[1][1] * q) % tt == 0;
      bool f2 = (A[1][1] * k - A[1][0] * q) % tt == 0;
      if (f1 && f2) good();
    } else {
      if (k == 0 && q == 0) good();
    }
  }
  printf("NO\n");
  return 0;
}
