可以发现结果就是求若干
可以用伯努利数快速计算,见:portal
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 60
#define mod 1000000007
inline char gc(){
static char buf[1<<16],*S,*T;
if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
return *S++;
}
inline ll read(){
ll x=0,f=1;char ch=gc();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
return x*f;
}
int inv[N],C[N][N],B[N];ll a[N];
inline int cal(int n,int k){
int res=0,tmp=n+1;
for(int i=1;i<=k+1;++i)
(res+=(ll)C[k+1][i]*B[k+1-i]%mod*tmp%mod)%=mod,tmp=(ll)tmp*(n+1)%mod;
res=(ll)res*inv[k+1]%mod;return res;
}
inline int ksm(int x,int k){
int res=1;for(;k;k>>=1,x=(ll)x*x%mod) if(k&1) res=(ll)res*x%mod;return res;
}
int main(){
// freopen("defile.in","r",stdin);
// freopen("defile.out","w",stdout);
int tst=read();inv[1]=1;B[0]=1;
for(int i=2;i<=52;++i) inv[i]=(ll)inv[mod%i]*(mod-mod/i)%mod;
for(int i=0;i<=52;++i) C[i][0]=1;
for(int i=1;i<=52;++i)
for(int j=1;j<=i;++j) C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
for(int i=1;i<=51;++i){
for(int j=0;j<=i-1;++j) (B[i]+=(ll)C[i+1][j]*B[j]%mod)%=mod;
B[i]=(ll)B[i]*inv[i+1]*-1%mod;if(B[i]<0) B[i]+=mod;
}while(tst--){
int n=read()%mod,m=read(),res=0;
for(int i=1;i<=m;++i) a[i]=read();sort(a+1,a+m+1);
for(int i=1;i<=m;++i) a[i]%=mod;
for(int i=0;i<=m;++i){
(res+=cal(n-a[i],m+1))%=mod;
for(int j=i+1;j<=m;++j)
(res-=ksm(a[j]-a[i],m+1))%=mod;
}printf("%d\n",res<0?res+mod:res);
}return 0;
}