#if 0

#endif

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef complex<double> xy_t;

double dot_product(xy_t a, xy_t b) {
  return (conj(a)*b).real();
}
double cross_product(xy_t a, xy_t b) {
  return (conj(a)*b).imag();
}
xy_t projection(xy_t p, xy_t b) {
  return b*dot_product(p,b)/norm(b);
}

xy_t P[110];
int main() {
  int N = 0;
  double x, y;
  while (scanf("%lf,%lf", &x, &y) != EOF)
    P[N++] = xy_t(x,y);
  double sum = 0.0;
  for (int i = 0; i + 2 < N; ++i) {
    xy_t a = P[0], b = P[i+1], c = P[i+2];

    sum += cross_product(b-a, c-a)/2.0;
  }
  printf("%.6f\n", abs(sum));
}


### Prompt

Please create a solution in CPP to the following problem:

Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.

However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.

<image>



input

The input is given in the following format:


x1, y1
x2, y2
::
xn, yn


xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.

output

Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.

Example

Input

0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0


Output

2.500000

### Response

```cpp
#if 0

#endif

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef complex<double> xy_t;

double dot_product(xy_t a, xy_t b) {
  return (conj(a)*b).real();
}
double cross_product(xy_t a, xy_t b) {
  return (conj(a)*b).imag();
}
xy_t projection(xy_t p, xy_t b) {
  return b*dot_product(p,b)/norm(b);
}

xy_t P[110];
int main() {
  int N = 0;
  double x, y;
  while (scanf("%lf,%lf", &x, &y) != EOF)
    P[N++] = xy_t(x,y);
  double sum = 0.0;
  for (int i = 0; i + 2 < N; ++i) {
    xy_t a = P[0], b = P[i+1], c = P[i+2];

    sum += cross_product(b-a, c-a)/2.0;
  }
  printf("%.6f\n", abs(sum));
}


```