Problem: Read a set of coordinates of three different points P1 (x1, y1), P2 (x2, y2), Q (xq, yq) on the plane, and line symmetry with point Q with the straight line passing through point P1 point P2 as the axis of symmetry. Create a program that outputs the point R (x, y) at the position of. Note that the point Q is not on its axis of symmetry.

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input

The input consists of multiple datasets. Each dataset is given in the following format:


x1, y1, x2, y2, xq, yq


x1, y1, x2, y2, xq, yq (real numbers between -100 and 100) are given on one line separated by commas.

The number of datasets does not exceed 50.

output

Print x, y on a single line, separated by blanks, for each dataset. The output may contain an error of 0.0001 or less in real numbers.

Example

Input

1.0,0.0,-1.0,0.0,1.0,1.0
1.0,0.0,0.0,-1.0,3.0,0.0
0.0,1.0,0.0,-1.0,1.0,1.0


Output

1.000000 -1.000000
1.000000 2.000000
-1.000000 1.000000

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <math.h>
#include <complex>
#include <cstdio>
#include <iomanip>
using namespace std;
#define rep(i,n) for (int i=0; i < (n); i++)

typedef double D;      // 座標値の型。doubleかlong doubleを想定
typedef complex<D> P;  // Point
typedef pair<P, P> L;  // Line
typedef vector<P> VP;
const D EPS = (1e-10);    // 許容誤差。問題によって変える
#define X real()
#define Y imag()
#define LE(n,m) ((n) < (m) + EPS)
#define GE(n,m) ((n) + EPS > (m))
#define EQ(n,m) (abs((n)-(m)) < EPS)

namespace std {
  bool operator<(const P a, const P b) {
    return a.X != b.X ? a.X < b.X : a.Y < b.Y;
  }
}

// 内積　dot(a,b) = |a||b|cosθ
D dot(P a, P b) {
  return (conj(a)*b).X;
}
// 外積　cross(a,b) = |a||b|sinθ
D cross(P a, P b) {
  return (conj(a)*b).Y;
}

int ccw(P a, P b, P c) {
  b -= a;  c -= a;
  if (cross(b,c) >  EPS) return +1;  // counter clockwise
  if (cross(b,c) < -EPS) return -1;  // clockwise
  if (dot(b,c)   < -EPS) return +2;  // c--a--b on line
  if (norm(b) < norm(c)) return -2;  // a--b--c on line or a==b
  return 0;                          // a--c--b on line or a==c or b==c
}

bool isecLP(P a1, P a2, P b) {
  return abs(ccw(a1, a2, b)) != 1;
}

bool isecLL(P a1, P a2, P b1, P b2) {
  return !isecLP(a2-a1, b2-b1, 0) || isecLP(a1, b1, b2);
}

VP convexHull(VP &ps) {
  int n = ps.size(), k = 0;
  sort(ps.begin(), ps.end());
  VP ch(2*n);
  for (int i = 0; i < n; ch[k++] = ps[i++]) // lower-hull
    while (k >= 2 && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;  // 余計な点も含むなら == -1 とする
  for (int i = n-2, t = k+1; i >= 0; ch[k++] = ps[i--]) // upper-hull
    while (k >= t && ccw(ch[k-2], ch[k-1], ps[i]) <= 0) --k;
  ch.resize(k-1);
  return ch;
}

D area(const VP& ps) {
  D a = 0;
  rep (i, ps.size()) a += cross(ps[i], ps[(i+1) % ps.size()]);
  return a / 2;
}

P proj(P a1, P a2, P p) {
  return a1 + dot(a2-a1, p-a1)/norm(a2-a1) * (a2-a1);
}

P reflection(P a1, P a2, P p) {
  return 2.0*proj(a1, a2, p) - p;
}

int main() {
  D x1, y1, x2, y2, xq, yq;
  while (scanf("%lf,%lf,%lf,%lf,%lf,%lf", &x1, &y1, &x2, &y2, &xq, &yq) != EOF) {
    P p1 = P(x1, y1);
    P p2 = P(x2, y2);
    P q = P(xq, yq);
    P ans = reflection(p1, p2, q);
    std::cout << fixed << setprecision(9) << ans.X << " " << ans.Y << std::endl;
  }
  return 0;
}