- 每次找l,r中出现次数最小的数,让他出现次数加一即可
- a[i],b[i]表示出现了a[i]次的数有b[i]个
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define ll long long
using namespace std;
const int mod=998244353;
const int N=2e6+1;
const int M=1e7+1;
int fact[N+M],b[N],a[N],T,n,m,l,r;
inline int read(){
int num=0;char ch=getchar();
while(!isdigit(ch))ch=getchar();
while(isdigit(ch))num=num*10+ch-'0',ch=getchar();
return num;
}
inline void Print(ll x){if(x>9)Print(x/10);putchar(x%10+'0');}
int power(int a,int b){int ans=1;for(;b;b>>=1,a=1LL*a*a%mod)if(b&1)ans=1LL*ans*a%mod;return ans;}
int main()
{
//freopen("a.in","r",stdin);
T=read();
fact[0]=1;rep(i,1,N+M-1)fact[i]=1LL*i*fact[i-1]%mod;
while(T--){
n=read(),m=read(),l=read(),r=read();
int f=r-l+1,cnt=0,tot=0,ans=1;
rep(i,1,n)a[i]=read();
sort(a+1,a+n+1);
for(int i=1,j=1;i<=n;i=j){
while(j<=n&&a[i]==a[j])j++;
if(a[i]<l||a[i]>r)ans=1LL*ans*fact[j-i]%mod;
else b[++cnt]=j-i,f--;
}
sort(b+1,b+cnt+1);
for(int i=1,j=1;i<=cnt;i=j){
while(j<=cnt&&b[i]==b[j])j++;
a[++tot]=b[i],b[tot]=j-i;
}
a[++tot]=1e9,b[tot]=1;
for(int i=1,M=m;i<=tot;i++){
int s=a[i]-a[i-1];
ll val=1LL*s*f;
if(val>M){
rep(j,i,tot-1)ans=1LL*ans*power(fact[a[j]],b[j])%mod;
int t=a[i-1]+M/f,last=M%f;
ans=1LL*ans*power(fact[t],f-last)%mod*power(fact[t+1],last)%mod;
break;
}else f+=b[i],M-=val;
}
Print(1LL*fact[n+m]*power(ans,mod-2)%mod);puts("");
}return 0;
}