Petr has just bought a new car. He's just arrived at the most known Petersburg's petrol station to refuel it when he suddenly discovered that the petrol tank is secured with a combination lock! The lock has a scale of 360 degrees and a pointer which initially points at zero:

<image>

Petr called his car dealer, who instructed him to rotate the lock's wheel exactly n times. The i-th rotation should be a_i degrees, either clockwise or counterclockwise, and after all n rotations the pointer should again point at zero.

This confused Petr a little bit as he isn't sure which rotations should be done clockwise and which should be done counterclockwise. As there are many possible ways of rotating the lock, help him and find out whether there exists at least one, such that after all n rotations the pointer will point at zero again.

Input

The first line contains one integer n (1 ≤ n ≤ 15) — the number of rotations.

Each of the following n lines contains one integer a_i (1 ≤ a_i ≤ 180) — the angle of the i-th rotation in degrees.

Output

If it is possible to do all the rotations so that the pointer will point at zero after all of them are performed, print a single word "YES". Otherwise, print "NO". Petr will probably buy a new car in this case.

You can print each letter in any case (upper or lower).

Examples

Input


3
10
20
30


Output


YES


Input


3
10
10
10


Output


NO


Input


3
120
120
120


Output


YES

Note

In the first example, we can achieve our goal by applying the first and the second rotation clockwise, and performing the third rotation counterclockwise.

In the second example, it's impossible to perform the rotations in order to make the pointer point at zero in the end.

In the third example, Petr can do all three rotations clockwise. In this case, the whole wheel will be rotated by 360 degrees clockwise and the pointer will point at zero again.

#include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ll long long
#define pb push_back
#define mp make_pair
#define endl "\n"
#define int ll
#define vi vector<int>
#define vb vector<bool>
#define vvb vector<vb >
#define pii pair<int,int>
#define ss second
#define ff first
#define vpii vector<pii>
#define vvi vector<vi >
#define vs vector<string>
#define vvs vector<vs >
#define pqi priority_queue <int>
#define minpqi priority_queue <int, vector<int>, greater<int> >
#define all(x) x.begin(),x.end()
#define mii map<int,int>
#define for0(i,n) for(ll i=0;i<n;i++)
#define for1(i,n) for(ll i=1;i<=n;i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define per1(i,n) for(ll i=n;i>0;i--)
#define repeat(i,start,n) for(ll i=start;i<n;i++)
#define inp(arr,n) ll arr[n];rep(i,n){ cin>>arr[i];}
#define inp1(arr,n) ll arr[n+1];rep1(i,n){ cin>>arr[i];}
#define inp2d(arr,n,m) ll arr[n][m];rep(i,n)rep(j,m)cin>>arr[i][j];
#define inpg(adj,m) rep(i,m){int a,b;cin>>a>>b;adj[a].pb(b);adj[b].pb(a);}
#define print(a,n) for(ll i=0;i<n;i++){ cout<<a[i]<<" ";}
#define print1(a,n) for(ll i=1;i<=n;i++){ cout<<a[i]<<endl;}
#define mod 1000000007
#define maxx 1000000000000000000
#define PI 3.141592653589793238462643383279
#define mmax(a,b,c) max(a,max(b,c))
#define mmin(a,b,c) min(a,min(b,c))
#define init(arr,a) memset(arr,a,sizeof(arr))
#define lb lower_bound
#define ub upper_bound
#define er equal_range
#define maxe *max_element
#define mine *min_element
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b){
	if(a.ss==b.ss)
		return a.ff<b.ff;
	return (a.second<b.second);
}
string convert_to_bit(int a,int bit){
	string s;
	while(a>0){
		s+=(a%2)+48;
		a/=2;
	}
	while(s.size()<bit)
		s+='0';
	reverse(s.begin(),s.end());
	return s;
}
int to_int(string s)
{
	int ans=0;
	string temp;
	int i=0;
	while(s[i]=='0')
	{
		i++;
	}
	if(i==s.size())
		return 0;
	temp.assign(s,i,s.size());
	int mul=1;
	for(i=temp.size()-1;i>=0;i--)
	{
		int a=temp[i]-'0';
		ans+=mul*a;
		mul*=10;
	}
	return ans;

}
string to_string(int n)
{
	string ans="";
	if(n==0)
		return "0";
	while(n>0)
	{
		int a=n%10;
		n/=10;
		char temp=a+'0';
		ans+=temp;
	}
	reverse(all(ans));
	return ans;
}
int bin_to_dec(int n)
{
	int num = n;
	int dec_value = 0;
	int base = 1;
	int temp = num;
	while (temp) {
		int last_digit = temp % 10;
		temp = temp / 10;
		dec_value += last_digit * base;
		base = base * 2;
	}
	return dec_value;
}
int binstr_to_dec(string s)
{
	int ans=0;
	for0(i,s.size())
		if(s[i]=='1')
			ans+=pow(2,s.size()-i-1);
	return ans;
}
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
void faltu()
{
	int a=2;
	while(a>0)a--;
}
int nCr(ll n, ll r)
{
	ll res = 1;
	if ( r > n - r )
		r = n-r;
	for (ll i = 0; i < r; ++i)
	{
		res *= (n - i);	//general term (n-x)/(1+x)
		res /= (i + 1);
	}
	return res;
}
//------------------------------------Code Starts here------------------------------------//
int n;
int a[200];
void solve()
{
	sort(a,a+n);
	int pre[n],suf[n];
	pre[0]=a[0];
	suf[n-1]=a[n-1];
	for(int i=1;i<n;i++)
	{
		pre[i]=pre[i-1]+a[i];
		suf[n-i-1]=suf[n-i]+a[n-i-1];
	}
	if(pre[n-1]==360){cout<<"YES"<<endl;return;}
	for0(i,n-1)
	{
		if(pre[i]==suf[i+1])
		{
			cout<<"YES";return;
		}
	}
	cout<<"NO"<<endl;
}
main()
{
  cin>>n;
  for0(i,n)cin>>a[i];
  solve();
}
