Problem: Long time ago Alex created an interesting problem about parallelogram. The input data for this problem contained four integer points on the Cartesian plane, that defined the set of vertices of some non-degenerate (positive area) parallelogram. Points not necessary were given in the order of clockwise or counterclockwise traversal.

Alex had very nice test for this problem, but is somehow happened that the last line of the input was lost and now he has only three out of four points of the original parallelogram. He remembers that test was so good that he asks you to restore it given only these three points.

Input

The input consists of three lines, each containing a pair of integer coordinates xi and yi ( - 1000 ≤ xi, yi ≤ 1000). It's guaranteed that these three points do not lie on the same line and no two of them coincide.

Output

First print integer k — the number of ways to add one new integer point such that the obtained set defines some parallelogram of positive area. There is no requirement for the points to be arranged in any special order (like traversal), they just define the set of vertices.

Then print k lines, each containing a pair of integer — possible coordinates of the fourth point.

Example

Input

0 0
1 0
0 1


Output

3
1 -1
-1 1
1 1

Note

If you need clarification of what parallelogram is, please check Wikipedia page:

https://en.wikipedia.org/wiki/Parallelogram

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long MOD_EXPO(long long b, long long p, long long m) {
  if (p == 0) return 1;
  long long ret = MOD_EXPO(b, p / 2, m) % m;
  ret = (ret * ret) % m;
  return ((p & 1) ? (ret * b) % m : ret % m);
}
long long POWER(long long N, long long K) {
  long long i, ans = 1;
  for (i = 1; i <= K; i++) ans *= N;
  return ans;
}
int SET(int N, int pos) { return (N | (1 << pos)); }
int RESET(int N, int pos) { return (N & !(1 << pos)); }
bool CHECK(int N, int pos) { return (N & (1 << pos)); }
int dx4[] = {1, -1, 0, 0};
int dy4[] = {0, 0, 1, -1};
int dx6[] = {0, 0, 1, -1, 0, 0};
int dy6[] = {1, -1, 0, 0, 0, 0};
int dz6[] = {0, 0, 0, 0, 1, -1};
int dx8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dy8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dkx8[] = {-1, 1, -1, 1, -2, -2, 2, 2};
int dky8[] = {2, 2, -2, -2, 1, -1, 1, -1};
int tc = 1;
const double eps = 1e-9;
const double pi = acos(-1.0);
const long long int mx = 1e5;
const long long int mod = 1e9 + 7;
int main() {
  int x[5], y[5], i, a, b, c, d, xx, yy;
  int X[5], Y[5];
  vector<pair<int, int> > vp;
  vector<pair<int, int> > vps;
  set<pair<int, int> > s;
  set<pair<int, int> >::iterator it;
  for (i = 0; i < 3; i++) {
    cin >> x[i] >> y[i];
    vp.push_back(pair<int, int>(y[i], x[i]));
  }
  cout << 3 << "\n";
  cout << x[0] - x[1] + x[2] << " " << y[0] - y[1] + y[2] << "\n";
  cout << x[0] + x[1] - x[2] << " " << y[0] + y[1] - y[2] << "\n";
  cout << -x[0] + x[1] + x[2] << " " << -y[0] + y[1] + y[2] << "\n";
  return 0;
}