human
CPP solution for "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":



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Give me a Python code to print "YES" (without the quotes) if it is possible to get vector B using the given operations?
assistant
#Here is Python code to print "YES" (without the quotes) if it is possible to get vector B using the given operations.
 #include <bits/stdc++.h>
using namespace std;
pair<int, int> transform(const pair<int, int>& org, int dir) {
  switch (dir) {
    case (0):
      return org;
    case (1):
      return make_pair(org.second, -org.first);
    case (2):
      return make_pair(-org.first, -org.second);
    case (3):
      return make_pair(-org.second, org.first);
  }
  assert(false);
}
int dir1[] = {0, 0, 2, 2};
int dir2[] = {1, 3, 1, 3};
void ok() {
  cout << "YES" << endl;
  exit(0);
}
long long GreatestCommonDivisor(long long abc, long long def) {
  if (abc < 0LL || def < 0LL) return GreatestCommonDivisor(abs(abc), abs(def));
  if (abc < def) return GreatestCommonDivisor(def, abc);
  if (!def) return abc;
  return GreatestCommonDivisor(def, abc % def);
}
int solve(long long x1, long long x2, long long x3) {
  return x3 % GreatestCommonDivisor(x1, x2) == 0LL;
}
int main() {
  pair<int, int> a;
  cin >> a.first >> a.second;
  pair<int, int> t;
  cin >> t.first >> t.second;
  pair<int, int> b;
  cin >> b.first >> b.second;
  for (int(i) = 0; (i) < (4); ++(i)) {
    pair<int, int> trans = transform(a, i);
    pair<int, int> target =
        make_pair(t.first - trans.first, t.second - trans.second);
    if (target.first == 0 && target.second == 0) {
      cout << "YES" << endl;
      return 0;
    }
    if (b.first == 0 && b.second == 0) {
      continue;
    }
    for (int(j) = 0; (j) < (4); ++(j)) {
      int d1 = dir1[j];
      int d2 = dir2[j];
      pair<int, int> trans1, trans2;
      trans1 = transform(b, d1);
      trans2 = transform(b, d2);
      long long x1 = trans1.first;
      long long y1 = trans1.second;
      long long x2 = trans2.first;
      long long y2 = trans2.second;
      long long x3 = target.first;
      long long y3 = target.second;
      if (x1 == 0LL && x2 == 0LL && x3 != 0LL) continue;
      if (y1 == 0LL && y2 == 0LL && y3 != 0LL) continue;
      if (x1 == 0LL && x2 == 0LL && x3 == 0LL) {
        if (solve(y1, y2, y3)) ok();
      } else if (y1 == 0LL && y2 == 0LL && y3 == 0LL) {
        if (solve(x1,...