LLink

显然的，答案就是$Cn^m*D{n-m}$

#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;
}