Problem: Arpa is taking a geometry exam. Here is the last problem of the exam.

You are given three points a, b, c.

Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.

Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.

Input

The only line contains six integers ax, ay, bx, by, cx, cy (|ax|, |ay|, |bx|, |by|, |cx|, |cy| ≤ 109). It's guaranteed that the points are distinct.

Output

Print "Yes" if the problem has a solution, "No" otherwise.

You can print each letter in any case (upper or lower).

Examples

Input

0 1 1 1 1 0


Output

Yes


Input

1 1 0 0 1000 1000


Output

No

Note

In the first sample test, rotate the page around (0.5, 0.5) by <image>.

In the second sample test, you can't find any solution.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long mod = pow(10, 9) + 7;
const double pi = 3.141592653589793238;
long long countdivisors(long long n) {
  long long count = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        count++;
      } else {
        count = count + 2;
      }
    }
  }
  return count;
}
void remove(std::vector<int> &v) {
  auto end = v.end();
  for (auto it = v.begin(); it != end; ++it) {
    end = std::remove(it + 1, end, *it);
  }
  v.erase(end, v.end());
}
bool isPerfectSquare(long long x) {
  long long s = sqrt(x);
  return (s * s == x);
}
bool isFibonacci(int n) {
  return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);
}
int digSum(int n) {
  int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
long long exponentMod(long long A, long long B, long long C) {
  if (A == 0) return 0;
  if (B == 0) return 1;
  long long y;
  if (B % 2 == 0) {
    y = exponentMod(A, B / 2, C);
    y = (y * y) % C;
  } else {
    y = A % C;
    y = (y * exponentMod(A, B - 1, C) % C) % C;
  }
  return (long long)((y + C) % C);
}
void sieve(int N) {
  bool isPrime[N + 1];
  for (int i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (int j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
void solve() {
  double xa, ya, xb, yb, xc, yc;
  cin >> xa >> ya >> xb >> yb >> xc >> yc;
  double m1 = (yb - ya) / (xb - xa);
  double m2 = (yc - yb) / (xc - xb);
  double m3 = (yc - ya) / (xc - xa);
  if (m1 == m2 && m1 == m3) {
    cout << "NO";
  } else {
    double d1 = sqrt(((ya - yb) * (ya - yb)) + ((xa - xb) * (xa - xb)));
    double d2 = sqrt(((yc - yb) * (yc - yb)) + ((xc - xb) * (xc - xb)));
    if (d1 == d2) {
      cout << "YES";
    } else {
      cout << "NO";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}