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;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& _p) {
  return os << "(" << _p.first << "," << _p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& _V) {
  bool f = true;
  os << "[";
  for (auto v : _V) {
    os << (f ? "" : ",") << v;
    f = false;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& _S) {
  bool f = true;
  os << "(";
  for (auto s : _S) {
    os << (f ? "" : ",") << s;
    f = false;
  }
  return os << ")";
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const map<T, U>& _M) {
  return os << set<pair<T, U>>(_M.begin(), _M.end());
}
const signed long long INF = 1000000100;
const long double EPS = 1e-9;
signed long long ax, ay, bx, by, cx, cy;
void read_data() { cin >> ax >> ay >> bx >> by >> cx >> cy; }
bool solve() {
  if (cx == 0 and cy == 0) {
    return (ax == bx and ay == by);
  }
  signed long long det = cx * cx + cy * cy;
  signed long long d1 = bx - ax;
  signed long long d2 = by - ay;
  signed long long det_p = d1 * cx + d2 * cy;
  signed long long det_q = d2 * cx - d1 * cy;
  return (det_p % det == 0) and (det_q % det == 0);
}
int main() {
  read_data();
  bool ok = false;
  for (int(_) = (1); (_) <= (4); (_)++) {
    signed long long nx = -ay;
    signed long long ny = ax;
    ax = nx;
    ay = ny;
    ok |= solve();
  }
  cout << (ok ? "YES" : "NO") << endl;
}
