#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <cstring>
#include <deque>
#include <cstdio>
#include <cmath>
#include <complex>
#include <iostream>
#include <iterator>

using namespace std;

#define reep(i,f,t) for(int i=f ; i<int(t) ; ++i)
#define rep(i,n) reep(i, 0, n) 

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;

const double EPS = 1e-8;

int main()
{
	int m;
	while(scanf("%d", &m), m){
		double sm1 = 0.0;
		int arg = 0;
		rep(i, m-1){
			int deg;
			scanf("%d", &deg);
			arg += deg;
			sm1 += sin(deg*M_PI/180.0);
		}
		sm1 += sin((360-arg)*M_PI/180.0);
		
		int n;
		scanf("%d", &n);
		
		double sm2 = 0.0; 
		arg = 0;
		rep(i, n-1){
			int deg;
			scanf("%d", &deg);
			arg += deg;
			sm2 += sin(deg*M_PI/180.0);
		}
		sm2 += sin((360-arg)*M_PI/180.0);
		
		if(fabs(sm1 - sm2) < EPS)
			puts("0");
		else
			puts(sm1>sm2 ? "1" : "2");
	}
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <vector>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <cstring>
#include <deque>
#include <cstdio>
#include <cmath>
#include <complex>
#include <iostream>
#include <iterator>

using namespace std;

#define reep(i,f,t) for(int i=f ; i<int(t) ; ++i)
#define rep(i,n) reep(i, 0, n) 

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;

const double EPS = 1e-8;

int main()
{
	int m;
	while(scanf("%d", &m), m){
		double sm1 = 0.0;
		int arg = 0;
		rep(i, m-1){
			int deg;
			scanf("%d", &deg);
			arg += deg;
			sm1 += sin(deg*M_PI/180.0);
		}
		sm1 += sin((360-arg)*M_PI/180.0);
		
		int n;
		scanf("%d", &n);
		
		double sm2 = 0.0; 
		arg = 0;
		rep(i, n-1){
			int deg;
			scanf("%d", &deg);
			arg += deg;
			sm2 += sin(deg*M_PI/180.0);
		}
		sm2 += sin((360-arg)*M_PI/180.0);
		
		if(fabs(sm1 - sm2) < EPS)
			puts("0");
		else
			puts(sm1>sm2 ? "1" : "2");
	}
}
```