#include <cstdio>
#include <complex>
#include <cmath>
using namespace std;

typedef complex<double> xy_t;

xy_t P[110];

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); }

int main(){
    int n = 0;
    double x, y;
    // "Control + d"??§?????¢
    while(~scanf("%lf, %lf", &x, &y)){
        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;
    }
    printf("%.6f\n", abs(sum));
}
### Prompt

Please formulate a Cpp solution 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
#include <cstdio>
#include <complex>
#include <cmath>
using namespace std;

typedef complex<double> xy_t;

xy_t P[110];

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); }

int main(){
    int n = 0;
    double x, y;
    // "Control + d"??§?????¢
    while(~scanf("%lf, %lf", &x, &y)){
        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;
    }
    printf("%.6f\n", abs(sum));
}
```