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;
long long ax, ay, bx, by, cx, cy;
int solve(long long x2, long long y2) {
  if (cx == 0 && cy == 0) {
    if (x2 == 0 && y2 == 0) return 1;
    return 0;
  } else if (cx == 0) {
    long long g, h;
    g = x2 / cy;
    h = y2 / cy;
    if (g * cy == x2 && h * cy == y2) return 1;
    return 0;
  } else if (cy == 0) {
    long long g, h;
    g = x2 / cx;
    h = y2 / cx;
    if (g * cx == x2 && h * cx == y2) return 1;
    return 0;
  } else {
    long long a, b;
    if ((cx * cy + cx * cx) == 0) return 0;
    b = (x2 * cy - y2 * cx) / (cy * cy + cx * cx);
    a = (x2 - b * cy) / cx;
    if (b * (cy * cy + cx * cx) == (x2 * cy - y2 * cx) &&
        (a * cx == x2 - b * cy)) {
      return 1;
    }
    return 0;
  }
  return 0;
}
int main() {
  scanf("%lld%lld", &ax, &ay);
  scanf("%lld%lld", &bx, &by);
  scanf("%lld%lld", &cx, &cy);
  if (solve(bx - ax, by - ay)) {
    puts("YES");
    return 0;
  }
  if (solve(bx + ax, by + ay)) {
    puts("YES");
    return 0;
  }
  if (solve(bx - ay, by + ax)) {
    puts("YES");
    return 0;
  }
  if (solve(bx + ay, by - ax)) {
    puts("YES");
    return 0;
  }
  puts("NO");
  return 0;
}
