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;
double const pi = 3.1415926535897932384626433832795;
int const inf = (int)1e9;
long long const inf64 = (long long)2e18;
const string name = "c";
long long ax, ay, bx, by, cx, cy;
void rotate(long long &x, long long &y) {
  long long xx = -y, yy = x;
  x = xx, y = yy;
}
void answer() {
  cout << "YES" << endl;
  exit(0);
}
void solve(long long x, long long y) {
  if (cx == 0 && cy == 0) {
    if (x == 0 && y == 0) answer();
    return;
  }
  if (cx == 0 || cy == 0) {
    long long zn = cx + cy;
    if (zn < 0) zn = -zn;
    if (x % zn == 0 && y % zn == 0) answer();
    return;
  }
  if ((x * cy - y * cx) % (cx * cx + cy * cy) == 0 &&
      (y * cy + x * cx) % (cx * cx + cy * cy) == 0)
    answer();
}
int main() {
  cin >> ax >> ay;
  cin >> bx >> by;
  cin >> cx >> cy;
  for (int i = 0; i < (int)4; i++) {
    for (int j = 0; j < (int)4; j++) {
      solve(bx - ax, by - ay);
      rotate(bx, by);
    }
    rotate(ax, ay);
  }
  cout << "NO" << endl;
  return 0;
}
