#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;int n,sum1,sum2,a[15];
int st1[1500],st2[1500],st[1500],num1,num2,num;
int total[1500],all;
//st1数组保存的是第一辆车能够一次运走的所有状态，同理st2数组为第二辆车的
//st数组保存的是两辆车一次能够运走的所有状态
void dfs(int f,int flag,int x,int s)//开始的时候这个dfs是作死的错，说明dfs写得还不够熟练
{int i,j,temp;if(flag==1&&f==n+1)st1[num1++]=x;else if(f==n+1)st2[num2++]=x;if(f>n) return ;temp=1<<(n-f);if(flag==1){if(s+a[f]<=sum1)dfs(f+1,flag,x+temp,s+a[f]);dfs(f+1,flag,x,s);}else{if(s+a[f]<=sum2)dfs(f+1,flag,x+temp,s+a[f]);dfs(f+1,flag,x,s);}
}void hebing()//两辆车的状态合并
{int i,j,t=2;int visited[1500],temp;memset(visited,0,sizeof(visited));for(i=1;i<num1;i++)for(j=1;j<num2;j++){temp=st1[i]|st2[j];if(visited[temp]==0){st[num++]=temp;visited[temp]=1;}}
}void solve(int T)//我这里是用组合解决的，虽然提交了之后发现用组合比用背包时间还少
{                //但是觉得可能是测试数据的原因，个人觉得还是背包靠谱些printf("Scenario #%d:\n",T);int temp[1500],t,res=0,x;int i,j,max=0,flag=0;int visited[1500];all=2;total[1]=0;for(i=1;i<=n;i++)max=max+(1<<(i-1));while(1){res++;t=1;memset(visited,0,sizeof(visited));for(i=1;i<all;i++){// int kao=0;//kao++;for(j=1;j<num;j++){x=(total[i]|st[j]);if(x==max){flag=1;break;}if(visited[x]==0){visited[x]=1;temp[t++]=x;}}if(flag==1)break;}if(flag==1)break;for(i=1;i<t;i++)total[i]=temp[i];all=t;}printf("%d\n",res);
}int main()
{int i,T,t;scanf("%d",&T);for(t=1;t<=T;t++){num1=1;num2=1;num=1;scanf("%d%d%d",&n,&sum1,&sum2);for(i=1; i<=n; i++)scanf("%d",&a[i]);dfs(1,1,0,0);dfs(1,2,0,0);hebing();solve(t);if(t!=T)printf("\n");}return 0;
}

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;int n,sum1,sum2,a[15];
int st1[1500],st2[1500],st[1500],num1,num2,num;
int total[1500],all;void dfs(int f,int flag,int x,int s)
{int i,j,temp;if(flag==1&&f==n+1)st1[num1++]=x;else if(f==n+1)st2[num2++]=x;if(f>n) return ;temp=1<<(n-f);if(flag==1){if(s+a[f]<=sum1)dfs(f+1,flag,x+temp,s+a[f]);dfs(f+1,flag,x,s);}else{if(s+a[f]<=sum2)dfs(f+1,flag,x+temp,s+a[f]);dfs(f+1,flag,x,s);}
}void hebing()
{int i,j,t=2;int visited[1500],temp;memset(visited,0,sizeof(visited));for(i=1;i<num1;i++)for(j=1;j<num2;j++){temp=st1[i]|st2[j];if(visited[temp]==0){st[num++]=temp;visited[temp]=1;}}
}int Min(int x,int y)
{return x>y?y:x;
}void solve(int T)
{printf("Scenario #%d:\n",T);int i,j,dp[1500];for(i=1;i<(1<<n);i++)dp[i]=100000000;dp[0]=0;for(i=1;i<num;i++){for(j=(1<<n)-1;j>=0;j--){if(dp[j]==100000000)continue;if((j&st[i])==0)dp[j|st[i]]=Min(dp[j|st[i]],dp[j]+1);}}printf("%d\n",dp[(1<<n)-1]);
}int main()
{int i,T,t;scanf("%d",&T);for(t=1;t<=T;t++){num1=1;num2=1;num=1;scanf("%d%d%d",&n,&sum1,&sum2);for(i=1; i<=n; i++)scanf("%d",&a[i]);dfs(1,1,0,0);dfs(1,2,0,0);hebing();solve(t);if(t!=T)printf("\n");}return 0;
}