#include <bits/stdc++.h>
using namespace std;
//define
#define int long long
#define UNIQUE(v) v.erase(unique(all(v)), v.end());
#define ZIP(v) sort(all(v)),UNIQUE(v)
#define ADD(a, b) a = (a + b) % mod
#define SUB(a, b) a = (a+mod-b)%mod
#define MUL(a, b) a = (a * b) % mod
#define repi(i,m,n) for(int i = m;i < n;i++)
#define drep(i,n,m) for(int i = n;i >= m;i--)
#define rep(i,n) repi(i,0,n)
#define rrep(i,n) repi(i,1,n+1)
#define chmin(x,y) x = min(x,y)
#define chmax(x,y) x = max(x,y)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(), v.rend()
#define dmp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define pf(x) push_front(x)
#define fi first
#define se second
// 許容する誤差ε
#define EPS (1e-10)
// 2つのスカラーが等しいかどうか
#define EQ(a,b) (abs((a)-(b)) < EPS)
// 2つのベクトルが等しいかどうか
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )
//template
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p){
  os << "(" << p.first << "," << p.second << ")";return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v){
  for (auto it = v.begin();it != v.end();++it){
    if(it != v.begin())os << " ";os << *it;
      }return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &mp){
  for(auto x: mp)os << "(" << x.first << "," << x.second << ")" << endl;
  return os;
}
template<typename T, int SIZE>
int array_length(const T (&)[SIZE]){return SIZE;}
template<typename T, int N>
void PRINTF(const T (&a)[N], int s = N, int t = -1, bool f = true){
	if(t == -1){rep(i,s){if(i)cout << " ";cout << a[i];}}
	else repi(i,s,t){if(i!=s)cout << " ";cout << a[i];}
	if(f)cout << "\n";
}
template<typename T, int N1, int N2>
void PRINTF(const T (&a)[N1][N2], int h = N1, int w = N2){
	rep(i,h){rep(j,w){cout << a[i][j] << " \n"[j==w-1];}}
}
//typedef
typedef complex<double> Point;
typedef pair<int,int> P;
typedef pair<int, P> PP;
typedef pair<P, int> Pi;
typedef vector<int> vi;
typedef deque<int> dq;
const int inf = 1e9+7;
const int INF = 1e18+7;



double cross(Point a, Point b) {
  return (a.real() * b.imag() - a.imag() * b.real());
}

int a[100], b[100];
signed main(){
  while(1){
    int n, m, s1 = 0, s2 = 0;
    scanf("%lld", &n);
    if(n == 0)break;
    rep(i,n-1){scanf("%lld", &a[i]);s1 += a[i];}
    scanf("%lld", &m);
    rep(i,m-1){scanf("%lld", &b[i]);s2 += b[i];}
    a[n-1] = 360-s1;b[m-1] = 360-s2;
    
    double S1 = 0, S2 = 0;
    rep(i,n){
      double r = a[i]/180.0*M_PI;
      Point p1 = Point(1, 0);
      Point p2 = Point(cos(r), sin(r));
      S1 += cross(p1, p2);
    }
    rep(i,m){
      double r = b[i]/180.0*M_PI;
      Point p1 = Point(1, 0);
      Point p2 = Point(cos(r), sin(r));
      S2 += cross(p1, p2);
    }
    if(EQ(S1, S2))printf("0\n");
    else if(S1-S2 > -EPS)printf("1\n");
    else printf("2\n");
  }
	return 0;
}










### Prompt

Construct a CPP code solution to the problem outlined:

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 <bits/stdc++.h>
using namespace std;
//define
#define int long long
#define UNIQUE(v) v.erase(unique(all(v)), v.end());
#define ZIP(v) sort(all(v)),UNIQUE(v)
#define ADD(a, b) a = (a + b) % mod
#define SUB(a, b) a = (a+mod-b)%mod
#define MUL(a, b) a = (a * b) % mod
#define repi(i,m,n) for(int i = m;i < n;i++)
#define drep(i,n,m) for(int i = n;i >= m;i--)
#define rep(i,n) repi(i,0,n)
#define rrep(i,n) repi(i,1,n+1)
#define chmin(x,y) x = min(x,y)
#define chmax(x,y) x = max(x,y)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(), v.rend()
#define dmp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define pf(x) push_front(x)
#define fi first
#define se second
// 許容する誤差ε
#define EPS (1e-10)
// 2つのスカラーが等しいかどうか
#define EQ(a,b) (abs((a)-(b)) < EPS)
// 2つのベクトルが等しいかどうか
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )
//template
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p){
  os << "(" << p.first << "," << p.second << ")";return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v){
  for (auto it = v.begin();it != v.end();++it){
    if(it != v.begin())os << " ";os << *it;
      }return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &mp){
  for(auto x: mp)os << "(" << x.first << "," << x.second << ")" << endl;
  return os;
}
template<typename T, int SIZE>
int array_length(const T (&)[SIZE]){return SIZE;}
template<typename T, int N>
void PRINTF(const T (&a)[N], int s = N, int t = -1, bool f = true){
	if(t == -1){rep(i,s){if(i)cout << " ";cout << a[i];}}
	else repi(i,s,t){if(i!=s)cout << " ";cout << a[i];}
	if(f)cout << "\n";
}
template<typename T, int N1, int N2>
void PRINTF(const T (&a)[N1][N2], int h = N1, int w = N2){
	rep(i,h){rep(j,w){cout << a[i][j] << " \n"[j==w-1];}}
}
//typedef
typedef complex<double> Point;
typedef pair<int,int> P;
typedef pair<int, P> PP;
typedef pair<P, int> Pi;
typedef vector<int> vi;
typedef deque<int> dq;
const int inf = 1e9+7;
const int INF = 1e18+7;



double cross(Point a, Point b) {
  return (a.real() * b.imag() - a.imag() * b.real());
}

int a[100], b[100];
signed main(){
  while(1){
    int n, m, s1 = 0, s2 = 0;
    scanf("%lld", &n);
    if(n == 0)break;
    rep(i,n-1){scanf("%lld", &a[i]);s1 += a[i];}
    scanf("%lld", &m);
    rep(i,m-1){scanf("%lld", &b[i]);s2 += b[i];}
    a[n-1] = 360-s1;b[m-1] = 360-s2;
    
    double S1 = 0, S2 = 0;
    rep(i,n){
      double r = a[i]/180.0*M_PI;
      Point p1 = Point(1, 0);
      Point p2 = Point(cos(r), sin(r));
      S1 += cross(p1, p2);
    }
    rep(i,m){
      double r = b[i]/180.0*M_PI;
      Point p1 = Point(1, 0);
      Point p2 = Point(cos(r), sin(r));
      S2 += cross(p1, p2);
    }
    if(EQ(S1, S2))printf("0\n");
    else if(S1-S2 > -EPS)printf("1\n");
    else printf("2\n");
  }
	return 0;
}










```