1 #include <cstdio>
2 #include <iostream>
3 #include <cstring>
4 #include <cmath>
5 #include <cctype>
6 #include <cstring>
7 #include <queue>
8 #include <algorithm>
9 using namespace std;
10
11 #define res register int
12 inline int read()
13 {
14 int x(0),f(1); char ch;
15 while(!isdigit(ch=getchar())) if(ch=='-') f=-1;
16 while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
17 return f*x;
18 }
19
20 inline int max(int x,int y){return x>y?x:y;}
21 inline double max(double x,double y){return x>y?x:y;}
22 inline int min(int x,int y){return x<y?x:y;}
23
24 const int N=50000+5;
25 int a[N],n,k,L,R,t;
26 int d[N],q[N],q1[N];
27 double s[N];//差分数组，前缀和数组
28
29 inline bool check(double m)
30 {
31 int l=1,r=0; //把i看成右端点（此处取得最大s），找s[i]最小的左端点
32 double ans(-1e9-7);
33 for(res i=1 ; i<=n ; i++)
34 {
35 if(i-L+1<=0) continue;
36 while(l<=r && s[q[r]]>=s[i-L+1]) r--;
37 //加入i-L+1(和之前比右端点向右挪了一位)
38 q[++r]=i-L+1;
39 while(i-q[l]+1>R && l <= r) l++;//排除非法
40 if(l<=r) ans=max(ans,s[i]-s[q[l]]);
41 }
42 return ans>m*k;
43 }
44
45 inline bool judge(double m)
46 {
47 for(res i=1 ; i<=n ; i++)
48 s[i]=s[i-1]+(double)d[i]-m;
49 return check(m);
50 }
51 int main()
52 {
53 freopen("average.in","r",stdin);
54 freopen("average.out","w",stdout);
55 int T=read();
56 while(T--)
57 {
58 double maxn(0.0);
59 n=read(); k=read(); L=read(); R=read();
60 for(res i=1 ; i<=n ; i++) a[i]=read();
61 for(res i=1 ; i<=n ; i++) d[i]=a[i]-a[i-1];
62 double l=0,r=1e9+7;
63 int head(1),tail(0),head1(1),tail1(0);
64 double ans(-100000000.0);
65 //特殊情况，取得最大值与最小值的两端点之间的距离小于L
66 //此处用线段树也可以
67 for(res i=1 ; i<=n ; i++)
68 {
69 while((head<=tail) && (a[q[tail]]>a[i])) tail--;
70 q[++tail]=i;
71 while((head1<=tail1) && (a[q1[tail1]]<a[i])) tail1--;
72 q1[++tail1]=i;
73 while((head<=tail) && i-q[head]+1 > L) head++;
74 while((head1<=tail1) &&i-q1[head1]+1 > L) head1++;
75 ans=max(ans,(double)(a[q1[head1]]-a[q[head]]) / (double)(L+k-1));
76 }
77 while(r-l>0.0000001)
78 {
79 double mid=(l+r)/2;
80 if(judge(mid)) l=mid,ans=max(ans,l);
81 else r=mid;
82 }
83 //翻转过来，再求一遍，即在刚才的右端点处取得最小s
84 reverse(a+1,a+n+1);
85 for(res i=1 ; i<=n ; i++) d[i]=a[i]-a[i-1];
86 l=0,r=1e9+7;
87 while(r-l>0.0000001)
88 {
89 double mid=(l+r)/2.0;
90 if(judge(mid)) l=mid,ans=max(ans,l);
91 else r=mid;
92 }
93 printf("%.4lf\n",ans);
94 }
95 return 0;
96 }
