#include <iostream>
#include <vector>
#include <cmath>
#define rep(i,n) for(int i=0;i<(n);i++)
#define PI 3.141592653589793238462643383279
#define EPS 10e-8

using namespace std;

int main()
{
	int m, n;
	int sum;
	vector <double> sa, sb;

	for (;; ){
		cin >> m;
		if (!m )
			break; 
		vector <int> a (m, 0 );
		sum = 0;
		rep (i, m - 1 ){
			cin >> a[i];
			sum += a[i];
		} // end rep
		a[m-1] = 360 - sum;
		cin >> n;
		vector <int> b (n, 0 );
		sum = 0;
		rep (j, n-1 ){
			cin >> b[j];
			sum += b[j];
		} // end rep
		b[n-1] = 360 - sum;
		double s = 0.0, t = 0.0;
		rep (i, m )
			s += sin ((double)a[i]*PI/180. );
		sa.push_back (s );
		rep (j, n )
			t += sin ((double)b[j]*PI/180. );
		sb.push_back (t );
	} // end for

	int size = sa.size();
	rep (i, size ){
//		sa[i] = fabs (sa[i] ); sb[i] = fabs (sb[i] );
		if (sa[i] > sb[i]+EPS ){
			cout << '1' << endl;
		}else if (sa[i]+EPS < sb[i] ){
			cout << '2' << endl;
		}else{
			cout << '0' << endl;
		} // end if 
	} // end for

	return 0;
}
### Prompt

Your challenge is to write a cpp solution to the following problem:

Create a program that inputs the vertex information of two polygons inscribed in one circle and outputs the magnitude relationship of their areas.

Assume that each vertex of the X-side is numbered counterclockwise from 1 to X (the figure shows an example when X = 4). However, the given polygon assumes that the center of the circle is contained inside, and the data on the position of vertex i is the angle v (1 ≤ v <180) of the central angle measured counterclockwise from vertex i to vertex i + 1. Integer). Regarding the magnitude relationship to be output, 1 (half-width number) when the area of ​​the first polygon is large, 2 (half-width number) when the area of ​​the second polygon is large, and when the areas are the same, Please output 0 (half-width number).

<image>



Input

A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:


m
v1
v2
::
vm −1
n
v1
v2
::
vn − 1


The first line gives the number of vertices of the first polygon m (3 ≤ m ≤ 50), and the following m − 1 line gives the information vi (1 ≤ vi <180) of the vertices i of the first polygon. ..

The following line gives the number of vertices n (3 ≤ n ≤ 50) of the second polygon, and the following n − 1 line gives the information vi (1 ≤ vi <180) for the vertices i of the second polygon.

The number of datasets does not exceed 200.

Output

Outputs the magnitude relationship (half-width numbers) for each data set on one line.

Example

Input

4
30
125
75
4
30
125
75
5
30
125
75
65
4
30
125
75
4
30
125
75
6
30
50
50
25
75
0


Output

0
1
2

### Response

```cpp
#include <iostream>
#include <vector>
#include <cmath>
#define rep(i,n) for(int i=0;i<(n);i++)
#define PI 3.141592653589793238462643383279
#define EPS 10e-8

using namespace std;

int main()
{
	int m, n;
	int sum;
	vector <double> sa, sb;

	for (;; ){
		cin >> m;
		if (!m )
			break; 
		vector <int> a (m, 0 );
		sum = 0;
		rep (i, m - 1 ){
			cin >> a[i];
			sum += a[i];
		} // end rep
		a[m-1] = 360 - sum;
		cin >> n;
		vector <int> b (n, 0 );
		sum = 0;
		rep (j, n-1 ){
			cin >> b[j];
			sum += b[j];
		} // end rep
		b[n-1] = 360 - sum;
		double s = 0.0, t = 0.0;
		rep (i, m )
			s += sin ((double)a[i]*PI/180. );
		sa.push_back (s );
		rep (j, n )
			t += sin ((double)b[j]*PI/180. );
		sb.push_back (t );
	} // end for

	int size = sa.size();
	rep (i, size ){
//		sa[i] = fabs (sa[i] ); sb[i] = fabs (sb[i] );
		if (sa[i] > sb[i]+EPS ){
			cout << '1' << endl;
		}else if (sa[i]+EPS < sb[i] ){
			cout << '2' << endl;
		}else{
			cout << '0' << endl;
		} // end if 
	} // end for

	return 0;
}
```