#include<bits/stdc++.h>
#define endl '\n'
#pragma GCC optimize("Ofast")
using namespace std;
const int maxn = 503;
bitset <250006> ans;
bitset <250006> tek;
bitset <250006> newtek;
int i,j,p,q,n,m,k,sum;
int a[maxn];
void rek(int l,int r,bitset <250006> tek)
{
if(l>r)return ;
if(l==r)
{
///managed to compute the knapsack in tek without this one
tek=(tek<<a[l]);
ans&=tek;
return ;
}
int mid = (l+r)/2;
///l mid; mid+1 r
newtek = tek;
for(i=r;i>=mid+1;i--)
newtek|=(newtek<<(2*a[i]));
rek(l,mid,newtek);
newtek = tek;
for(i=l;i<=mid;i++)
newtek|=(newtek<<(2*a[i]));
rek(mid+1,r,newtek);
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin>>n;
for(i=1; i<=n; i++)
{
cin>>a[i];
sum+=a[i];
}
if(sum%2!=0)
{
cout<<0<<endl;
return 0;
}
for(i=2;i<=n;i++)
{
if(a[i]%2!=a[i-1]%2)
{
cout<<0<<endl;
return 0;
}
}
newtek[0]=1;
for(i=1;i<=n;i++)
newtek|=(newtek<<(a[i]));
if(!newtek[sum/2])
{
cout<<0<<endl;
return 0;
}
ans.set();
tek.set(0);
tek[0]=1;
rek(1,n,tek);
int br = 0;
for(i=1;i<sum;i++)
if(ans[i])br++;
cout<<br<<endl;
for(i=sum-1;i>1;i--)
{
if(ans[i])
cout<<sum-i<<" ";
}
return 0;
}
/**
4
1 3 1 5
4
200 200 200 200
*/
