#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 5e5+10;
const double eps = 1e-6;
double a[maxn];
double c[maxn];
int head1,head2,tail1,tail2;
int q1[maxn],q2[maxn];
int n,k,L,R;
int q3[maxn];
inline int read(){//快读
	int s = 0,f = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9'){
		if(ch == '-')f = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9'){
		s = s * 10 + ch - '0';
		ch = getchar();
	}
	return s * f;
}
bool jud(double mid){//二分判断函数
	memset(c,0,sizeof(c));
	for(int i=1;i<=n;++i){//记录第一种相减的情况
		c[i] = a[i] - i * mid;
	}
	double ans = -100000.0;
	int head3=0,tail3=0;
	for(int i=L+1;i<=n;++i){//单调队列维护最大值
		while(head3 < tail3 && i - q3[head3] >= R)head3++;
		while(head3 < tail3 && c[q3[tail3-1]] >= c[i-L])tail3--;
		q3[tail3++] = i-L;//记录最小值的位置
		ans = max(ans,c[i] - c[q3[head3]]);
	}
	head3=0,tail3=0;
	for(int i=1;i<=n;++i){//找另一种相加的情况
		c[i] = a[i] + i * mid;
	}
	for(int i=n-L;i;--i){//单调队列维护最大值
		while(head3 < tail3 && i - q3[head3] >= R)head3++;
		while(head3 < tail3 && c[q3[tail3-1]] >= c[i+L])tail3--;
		q3[tail3++] = i+L;
		ans = max(ans,c[i] - c[q3[head3]]);
	}
	return ans - mid * k >= 0;//判断是否成立
}
double calc1(){//计算长度为L的情况下的答案
	double ans = -100000.0;
	head1 = head2 = tail1 = tail2 = 0;//一定要初始化，不然爆掉
	for(int i=1;i<=n;++i){
		if(i < L){//位置不超过L的时候只维护单调性
			while(head1 < tail1 && a[q1[tail1-1]] >= a[i])tail1--;
			while(head2 < tail2 && a[q2[tail2-1]] <= a[i])tail2--;
		}
		else {//位置超过L的时候维护单调性且长度超过L的时候弹出队首
			while(head1 < tail1 && i - q1[head1] >= L)head1++;
			while(head2 < tail2 && i - q2[head2] >= L)head2++;
			while(head1 < tail1 && a[q1[tail1-1]] >= a[i])tail1--;
			while(head2 < tail2 && a[q2[tail2-1]] <= a[i])tail2--;
		}
		q1[tail1++] = i;
		q2[tail2++] = i;
		if(i >= L)//算答案
			ans = max(ans,(double)(a[q2[head2]] - a[q1[head1]]) / (L-1+k));
	}
	return ans;
}
int main(){
	int T = read();
	while(T--){
		n = read();
		k = read();
		L = read();
		R = read();
		for(int i=1;i<=n;++i){
			scanf("%lf",&a[i]);
		}
		double ans = calc1();//长度刚好为L的答案
		double l = 0,r = 1000000.0;
		while(r - l >= eps){//二分找到最优解
			double mid = (l + r) / 2;
			if(jud(mid))l = mid;
			else r = mid;
		}
		printf("%.4lf\n",max(ans,l));//取两种情况的最大值
	}
	return 0;
}