• We make a $$$sparse\ table$$$ containing where to go at $$$2^i$$$ th jump.
• We make another $$$sparse\ table$$$ containing the $$$range\ AND$$$ value that we will get at $$$2^i$$$ th jump.
• Now for each index, we find the maximum steps so that the range $$$AND$$$ remains $$$1$$$. Let us denote the steps as $$$len$$$. If it remains $$$1$$$ forever then the answer is $$$-1$$$. Otherwise we do it for all indices and take the $$$MAX_{1\le i\le n}\ len_i+1$$$, as $$$AND$$$ value shows a $$$monotonic$$$ property [the property of binary search]. That will be our desired answer.

Time Complexity: $$$O(n\log{n})$$$

130189104

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
 
//mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());	
 
ll ara[505];
ll dp[505][505];
 
int n;
 
ll f(int now,int nisi,vector<int>&vec)
{
	if(now>n)
	{
		ll ans=0;
		for(int i=0;i<61;i++)
		{
			if(vec[i]>=max(1,nisi-2)) ans+=(1LL<<i);
		}
		return ans;
	}
	
	ll &ans=dp[now][nisi];
	if(ans!=-1) return ans;
	
	vector<int>temp(vec.begin(),vec.end());
	
	for(int i=0;i<61;i++) if(ara[now]&(1LL<<i)) temp[i]++;
	
	ans=max(f(now+1,nisi+1,temp),f(now+1,nisi,vec));
	
	return ans;
}
 
void solve()
{
	cin>>n;
	
	for(int i=1;i<=n;i++) cin>>ara[i];
	
	memset(dp,-1,sizeof dp);
	vector<int>vec(61,0);
	
	cout<<f(1,0,vec);
}		
 
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	
	solve();
	
	#ifdef LOCAL
	cerr<<"\nTime elapsed: " << 1000.0 * clock() / CLOCKS_PER_SEC << " ms\n";
	#endif
	
	
	return 0;
}