/*
     http://hi.baidu.com/forverlin1204/blog/item/96eeab102a2a6dcda6ef3f61.html
    状态压缩DP 
    给定两辆车的容量c1,c2,物品的个数n，每件物品的重量w[i]，要使来回趟数尽可能少
    枚举1<<n种组合，看哪些物品可以由两辆车一次运走，然后DP求至少要运的次数 
    state存储可以一次由两辆车运完的状态 
 */
#include < cstdio >
#include < cstring >

int  n,c1,c2;
 int  w[ 12 ],v[ 12 ];
 int  dp[ 1 << 11 ];
 int  state[ 1 << 11 ],tot;

 // 选中k个，再用2进制组合
void  comb(){
     int  limit = 1 << n;
    tot = 0 ;
     for ( int  i = 1 ;i < limit;i ++ ){
         int  k = 0 ;
         for ( int  t = 0 ;t < n;t ++ ) if (i & ( 1 << t))v[k ++ ] = w[t];
         int  klimit = 1 << k;
         for ( int  j = 0 ;j < klimit;j ++ ){ // 组合 check， 1谁取，0谁取，注意j=0开始
             int  t1 = c1,t2 = c2;
             for ( int  t = 0 ;t < k;t ++ ){
                 if (j & ( 1 << t))t1 -= v[t];
                 else  t2 -= v[t];
                 if (t1 < 0 || t2 < 0 ) break ;
            }
             if (t1 >= 0 && t2 >= 0 ){state[ ++ tot] = i; break ;}
        }
    }
}
 void  DP(){ // 0-1
    memset(dp, - 1 , sizeof (dp));
    dp[ 0 ] = 0 ;
     for ( int  i = 1 ;i <= tot;i ++ )
         for ( int  j = ( 1 << n) - 1 - state[i];j >= 0 ;j -- ){
             if (dp[j] < 0 ) continue ;
             int  k = j + state[i];
             if ((j | state[i]) != k) continue ; //
              if (dp[k] ==- 1 || dp[k] > dp[j] + 1 )dp[k] = dp[j] + 1 ;
        }
}
 int  main(){
     int  T,t = 1 ;
    scanf( " %d " , & T);
     while (T -- ){
        scanf( " %d%d%d " , & n, & c1, & c2);
         for ( int  i = 0 ;i < n;i ++ )
            scanf( " %d " , & w[i]);
        comb();
        DP();
        printf( " Scenario #%d:\n%d\n\n " ,t ++ ,dp[( 1 << n) - 1 ]);
    }
     return   0 ;
}