Description

Input

Output

輸出 T 行，每行一個數，表示求出的序列數

Sample Input

Sample Output

HINT

Source

鳴謝Menci上傳

/*
水水的組合數+錯排 
C(n,m)*D[n-m]
tm bzoj為什麼還是CE!! 
 
*/
#include<bits/stdc++.h>

#define N 1000002
#define M 1000000007

using namespace std;
int n,m;
long long inv[N]={1,1},fac[N]={1,1},f[N]={1,1},D[N];

inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='
 
9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

inline int C(int a,int b)
{
    return ((fac[a]*inv[b])%M*inv[a-b]%M)%M;
}

inline void init()
{
    fac[1]=1;
    for(int i=2;i<=N;i++)
    {
        fac[i]=(1ll*fac[i-1]%M*i)%M;
        f[i]=((M-M/i)*f[M%i])%M;
        inv[i]=(inv[i-1]*f[i])%M;
    } 
}

int main()
{
    //freopen("ly.in","r",stdin);
    D[0]=1;D[2]=1;init();
    for(int i=3;i<=N;i++)
    D[i]=((i-1)*(D[i-1]%M+D[i-2]%M))%M;
    int T;cin>>T;
    while(T--)
    {
        n=read();m=read();
        if (n-m==1) printf("0\n");
        else if(m==n) printf("1\n");
        else if(m==0) cout<<D[n]<<endl;
        else  cout<<(1ll*C(n,m)*D[n-m])%M<<endl;
    }
    return 0;
}