//02
#include<iostream>
#include<cmath>

using namespace std;

int main(){
  for(int m;cin>>m,m;){
    int mr=360;
    double ms=0;
    for(int i=0;i<m-1;i++){
      int a;
      cin>>a;
      mr-=a;
      ms+=sin(acos(-1)*a/180);
    }
    ms+=sin(acos(-1)*mr/180);
    int nr=360;
    double ns=0;
    int n;
    cin>>n;
    for(int i=0;i<n-1;i++){
      int a;
      cin>>a;
      nr-=a;
      ns+=sin(acos(-1)*a/180);
    }
    ns+=sin(acos(-1)*nr/180);
    int ans;
    if(ms>ns+1e-9){
      ans=1;
    }else if(ms+1e-9<ns){
      ans=2;
    }else{
      ans=0;
    }
    cout<<ans<<endl;
  }
  return 0;
}
    
### Prompt

Please formulate 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
//02
#include<iostream>
#include<cmath>

using namespace std;

int main(){
  for(int m;cin>>m,m;){
    int mr=360;
    double ms=0;
    for(int i=0;i<m-1;i++){
      int a;
      cin>>a;
      mr-=a;
      ms+=sin(acos(-1)*a/180);
    }
    ms+=sin(acos(-1)*mr/180);
    int nr=360;
    double ns=0;
    int n;
    cin>>n;
    for(int i=0;i<n-1;i++){
      int a;
      cin>>a;
      nr-=a;
      ns+=sin(acos(-1)*a/180);
    }
    ns+=sin(acos(-1)*nr/180);
    int ans;
    if(ms>ns+1e-9){
      ans=1;
    }else if(ms+1e-9<ns){
      ans=2;
    }else{
      ans=0;
    }
    cout<<ans<<endl;
  }
  return 0;
}
    
```