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.

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

#define N 100 //頂点の最大個数

int main(void) {
	double x[N], y[N];
	int n; //頂点の個数
	double S; //面積
	
	n = 0;
	while(scanf("%lf,%lf", &x[n], &y[n]) != EOF) ++n;
	
	//座標(x[0],y[0])を含む三角形の面積を求める。
	S = 0;
	for(int i = 2; i < n; ++i) {
		int p1 = i - 1;
		int p2 = i;
		double a, b, c, z;
		a = sqrt((x[p1]-x[0])*(x[p1]-x[0]) + (y[p1]-y[0])*(y[p1]-y[0]));
		b = sqrt((x[p2]-x[0])*(x[p2]-x[0]) + (y[p2]-y[0])*(y[p2]-y[0]));
		c = sqrt((x[p1]-x[p2])*(x[p1]-x[p2]) + (y[p1]-y[p2])*(y[p1]-y[p2]));
		z = (a + b + c) / 2;
		S += sqrt(z * (z-a) * (z-b) * (z-c));
	}
	
	printf("%.6lf\n", S);
	
	return 0;
}
