#include<bits/stdc++.h>
#define int long long
using namespace std;
void ts(){cout<<"IAKIOI\n";}
inline int read(){
	int n=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		n=n*10+ch-'0';
		ch=getchar();
	}
	return n*f;
}
int n,m;
int a[1000005];
int fac[1000005],ifac[1000005];
const int mod=998244353;
int qkp(int b,int p){
	int r=1;
	while(p){
		if(p&1)(r*=b)%=mod;
		p/=2;
		(b*=b)%=mod;
	}
	return r;
}
int C(int n,int m){
	if(n<m)return 0;
	return fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}
void solve(){
	n=read(),m=read();
	for(int i=1;i<=m;i++)a[i]=read();
	bool h=0;
	for(int i=1;i<=m;i++)h|=(a[i]<=1);
	if(h){
		//(n-m)!*C(n,m)
		cout<<fac[n-m]*C(n,m)%mod<<"\n";
		return;
	}
	//没有 0 或 1.
	//怎么办? 
	//最后一步肯定是找
	int mn=1ll<<60;
	for(int i=1;i<=m;i++)mn=min(mn,a[i]);
	//0~mn-1 得是在一起的.
	//0~2.
	//n-m-(mn-1)+1.
	int v=fac[mn]*fac[n-m-(mn-1)]%mod;
	v*=C(n-m-mn+2+m-1,m);
	v%=mod;
	cout<<v<<"\n";
}
signed main(){
	fac[0]=1;
	for(int i=1;i<=1000004;i++)fac[i]=fac[i-1]*i%mod;
	ifac[1000004]=qkp(fac[1000004],mod-2);
	for(int i=1000003;i>=0;i--)ifac[i]=ifac[i+1]*(i+1)%mod;
	int t=read();
	while(t--)solve();
	return 0;
}
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