/*
IZhO 17-bootfall
Do knapsack for the initial n items. We will store count[i] = how many subsets have sum i, mod some big prime.
When we want to remove 1 item, we have to reverse the knapsack and then revert back to the initial state.
At the end, we need to check how many differences appear in all of the n removals, and this can be counted with a frequency array.
*/
#include<bits/stdc++.h>
#define god dimasi5eks
#pragma GCC optimize("O3")
#define fi first
#define se second
#define pb push_back
#define pf push_front
#define mod 1000000007
#define dancila 3.14159265359
#define eps 1e-9
using namespace std;
typedef long long ll;
int add(int a, int b)
{
ll x = a+b;
if(x >= mod)
x -= mod;
if(x < 0)
x += mod;
return x;
}
ll mul(ll a, ll b)
{
return (a*b) % mod;
}
ll pw(ll a, ll b)
{
ll ans = 1;
while(b)
{
if(b & 1)
ans = (ans * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand_seed()
{
long long a = rng();
return a;
}
int n;
int v[502];
int frq[1000002];
bool dp[250002];
int ct[250002], ct2[250002];
int fi[250002];
int lst[250002];
bool canruk(int lack)
{
int sm = 0;
for(int i = 1; i <= n; ++i)
if(i != lack)
sm += v[i];
if(lack == 0)
{
int mxval = 0;
dp[0] = 1;
ct[0] = 1;
for(int i = 1; i <= n; ++i)
{
for(int j = mxval; j >= 0; --j)
if(dp[j])
{
ct[j + v[i]] = add(ct[j + v[i]], ct[j]);
if(!fi[j + v[i]])
fi[j + v[i]] = i;
lst[j + v[i]] = i;
dp[j + v[i]] = 1;
}
mxval += v[i];
}
if(mxval % 2 == 0 && dp[mxval / 2])
return 1;
return 0;
}
else
{
for(int j = 0; j <= sm + v[lack]; ++j)
ct2[j] = ct[j];
for(int j = v[lack]; j <= sm + v[lack]; ++j)
ct[j] = add(ct[j], -ct[j - v[lack]]);
for(int i = 0; i <= sm/2; ++i)
if(ct[i])
frq[sm - 2 * i]++;
for(int j = 0; j <= sm + v[lack]; ++j)
ct[j] = ct2[j];
return 1;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n;
for(int i = 1; i <= n; ++i)
cin >> v[i];
sort(v+1, v+n+1);
if(!canruk(0))
{
cout << 0;
return 0;
}
for(int i = 1; i <= n; ++i)
canruk(i);
int nrok = 0;
for(int i = 0; i <= 1000000; ++i)
if(frq[i] == n)
++nrok;
cout << nrok << '\n';
for(int i = 0; i <= 1000000; ++i)
if(frq[i] == n)
cout << i << " ";
return 0;
}
