4
1024 512 256 256
4
1024 1024 1024 1024
5
1024 512 512 512 1
0

Case #1: 1
Case #2: 11
Case #3: 8

Hint
In the first case, we should choose all the numbers.
In the second case, all the subsequences which contain more than one number are good.
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#include <sstream>
#define maxn 100005
#define MAXN 100005
#define mod 998244353
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-8
typedef long long ll;
using namespace std;

int n,m,tot;
int mp[2050],cnt[13],p[13],ed[13];
ll ans,dp[13][2050];
ll inv[100005];
int a[12]= {2048,1024,512,256,128,64,32,16,8,4,2,1};

void scanf_(int&ret)
{
    char c;
    ret=0;
    while((c=getchar())<'0'||c>'9');
    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
}
ll pow_mod(ll x,ll n)
{
    ll res = 1;
    while(n)
    {
        if(n&1) res = res * x %mod;
        x = x * x %mod;
        n >>= 1;
    }
    return res;
}
void egcd(ll a,ll b,ll &x,ll &y)
{
    if (b==0)
    {
        x=1,y=0;
        return ;
    }
    egcd(b,a%b,x,y);
    ll t=x;
    x=y;
    y=t-a/b*x;
}
void solve()
{
    int i,j,k;
    memset(dp,0,sizeof(dp));
    for(i=0; i<=11; i++)
    {
        ed[i]=min(cnt[i],a[i]-1);
    }
    dp[0][0]=1;
    ll h=1;
    for(i=1; i<=ed[0]; i++)
    {
        h=((h*(cnt[0]-i+1))%mod)*inv[i]%mod;
        dp[0][i]=h;
    }
    for(i=0; i<11; i++)
    {
        for(j=0; j<a[i]; j++)
        {
            if(!dp[i][j]) continue ;
            h=1;
            dp[i+1][j>>1]+=dp[i][j];
            dp[i+1][j>>1]%=mod;
            for(k=1; k<=ed[i+1]; k++)
            {
                if((j>>1)+k<a[i+1])
                {
                    h=((h*(cnt[i+1]-k+1))%mod)*inv[k]%mod;
                    dp[i+1][(j>>1)+k]+=h*dp[i][j];
                    dp[i+1][(j>>1)+k]%=mod;
                }
                else break ;
            }
        }
    }
    ans=((pow_mod(2,n-tot)-dp[11][0]+mod)%mod)*pow_mod(2,tot);
    ans%=mod;
}
int main()
{
    int i,j,u,test=0;
    for(i=1; i<=100000; i++)
    {
        ll x,y;
        egcd(i,mod,x,y);
        x=(x+mod)%mod;
        inv[i]=x;
    }
    memset(mp,-1,sizeof(mp));
    p[0]=1;
    mp[1]=0;
    for(i=1; i<=11; i++)
    {
        p[i]=p[i-1]*2;
        mp[p[i]]=i;
    }
    while(1)
    {
        scanf_(n);
        if(n==0) break ;
        tot=0;
        memset(cnt,0,sizeof(cnt));
        for(i=1; i<=n; i++)
        {
            scanf_(u);
            if(mp[u]!=-1) cnt[mp[u]]++;
            else tot++;
        }
        solve();
        printf("Case #%d: %I64d\n",++test,ans);
    }
    return 0;
}
/*
11
256 256 256 256 256 256 512 512 1024 1024 2048
*/