显然的，答案就是\(C_n^m*D_{n-m}\)
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<ctime>
#include<bitset>
using namespace std;
int t;
long long n,m;
const long long mod=1e9+7;
long long f[1000005];
long long inv[1000005];
long long d[1000005];
long long qm(long long a,long long b){
long long res=1;
while(b>0){
if(b&1) res=(a*res)%mod;
a*=a;
a%=mod;
b>>=1;
}
return res;
}
int main(){
scanf("%d",&t);
f[0]=1;
inv[0]=1;
for(long long i=1;i<=1000002;++i){
f[i]=(f[i-1]*i)%mod;
inv[i]=qm(f[i],mod-2);
}
d[1]=0;
d[2]=1;
d[3]=2;
for(long long i=4;i<=1000002;++i){
d[i]=(i-1)*(d[i-1]+d[i-2])%mod;
}
while(t--){
scanf("%lld%lld",&n,&m);
if(m==0) {
printf("%lld\n",(d[n]+mod)%mod);
}else if(m!=n){
printf("%lld\n",(f[n]*inv[m]%mod*inv[n-m]%mod*d[n-m]%mod+mod)%mod);
}else{
cout<<1<<endl;
}
}
return 0;
}