#include <iostream>
#include <vector>
#define _USE_MATH_DEFINES
#include <cmath>

using namespace std;

void solve()
{
	int m;
	while(cin >>m, m)
	{
		double angle_remain1 = 360;
		double angle1;
		double sum1 = 0;
		for(int i = 1; i < m; ++i)
		{
			cin >> angle1;
			angle_remain1 -= angle1;
			sum1 += sin(angle1 / 180 * M_PI);
		}
		sum1 += sin(angle_remain1 / 180 * M_PI);
		int n;
		cin >> n;
		double angle_remain2 = 360;
		double angle2;
		double sum2 = 0;
		for(int i = 1; i < n; ++i)
		{
			cin >> angle2;
			angle_remain2 -= angle2;
			sum2 += sin(angle2 / 180 * M_PI);
		}
		sum2 += sin(angle_remain2 / 180 * M_PI);
		if(fabs(sum1 - sum2) < 0.000000001)
		{
			cout << 0 << endl;
		}
		else if(sum1 > sum2)
		{
			cout << 1 << endl;
		}
		else if(sum1 < sum2)
		{
			cout << 2 << endl;
		}
	}
}

int main()
{
	solve();
	return(0);
}
### Prompt

Generate 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>
#define _USE_MATH_DEFINES
#include <cmath>

using namespace std;

void solve()
{
	int m;
	while(cin >>m, m)
	{
		double angle_remain1 = 360;
		double angle1;
		double sum1 = 0;
		for(int i = 1; i < m; ++i)
		{
			cin >> angle1;
			angle_remain1 -= angle1;
			sum1 += sin(angle1 / 180 * M_PI);
		}
		sum1 += sin(angle_remain1 / 180 * M_PI);
		int n;
		cin >> n;
		double angle_remain2 = 360;
		double angle2;
		double sum2 = 0;
		for(int i = 1; i < n; ++i)
		{
			cin >> angle2;
			angle_remain2 -= angle2;
			sum2 += sin(angle2 / 180 * M_PI);
		}
		sum2 += sin(angle_remain2 / 180 * M_PI);
		if(fabs(sum1 - sum2) < 0.000000001)
		{
			cout << 0 << endl;
		}
		else if(sum1 > sum2)
		{
			cout << 1 << endl;
		}
		else if(sum1 < sum2)
		{
			cout << 2 << endl;
		}
	}
}

int main()
{
	solve();
	return(0);
}
```