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;
int ax, ay, bx, by, cx, cy;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int main(void) {
  scanf("%d%d%d%d%d%d", &ax, &ay, &bx, &by, &cx, &cy);
  if (cx == 0 && cy == 0) {
    if ((ax == bx && ay == by) || (ax == -by && ay == bx) ||
        (ax == -bx && ay == -by) || (ax == by && ay == -bx))
      puts("YES");
    else
      puts("NO");
    return 0;
  }
  for (int i = 0; i < 4; i++) {
    int rx = bx + dx[i] * ax + dy[i] * ay, ry = by + dx[i] * ay - dy[i] * ax;
    long long D = -cy * 1LL * cy - cx * 1LL * cx;
    long long DX = rx * 1LL * -cx - cy * 1LL * ry;
    long long DY = cx * 1LL * ry - rx * 1LL * cy;
    if (DX % D == 0 && DY % D == 0) {
      puts("YES");
      return 0;
    }
  }
  puts("NO");
  return 0;
}
