/*C(n,k)*/#include
     
      #include
      
       #include
       
        #define ll long long#define N 1000007#define M 1000000007using namespace std;ll n,k,ans,cnt1,cnt0;ll inv[N]={
   1,1},fac[N]={
   1,1},f[N]={
   1,1};inline ll 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 void init(){    for(int i=2;i<=N;i++)    {        fac[i]=(fac[i-1]%M*i%M)%M;        f[i]=((M-M/i)%M*f[M%i])%M;        inv[i]=(inv[i-1]*f[i]%M)%M;    }}inline ll C(ll a,ll b){    return fac[a]%M*inv[b]%M*inv[a-b]%M;}int main(){    freopen("cube.in","r",stdin);    freopen("cube.out","w",stdout);    n=read();k=read();init();    printf("%I64d\n",C(n,k));    return 0;}
       
      
     

 

 

 

/*做最大生成树时标记哪些边被加了进去。然后把这些边排序。如果询问权值w的货物，就在边中查找小于w的最大值假设拍第k大答案就是k+1。 复杂度O(qlogn) */#include
     
      #include
      
       #include
       
        #include
        
          #define N 100007using namespace std;int n,m,q,ans,cnt,num;int head[N],fa[N];struct edge{    int u,v,w,net,pos;    bool operator < (const edge &a) const{            return w>a.w;    }}e[N<<1],E[N],tmp[N<<1];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 void add(int u,int v,int w){    e[++cnt].u=u;e[cnt].v=v;e[cnt].w=w;    e[cnt].net=head[u];head[u]=cnt;}int find(int x){
   return fa[x]==x?x:fa[x]=find(fa[x]);}void kruskal(){    int r1,r2;    sort(E+1,E+m+1);    for(int i=1;i<=m;i++)    {        r1=find(E[i].u),r2=find(E[i].v);        if(r1==r2)continue;        fa[r1]=r2;num++;        add(E[i].u,E[i].v,E[i].w);add(E[i].v,E[i].u,E[i].w);        tmp[num].pos=num;tmp[num].u=E[i].u,tmp[num].v=E[i].v;tmp[num].w=E[i].w;        if(num==n-1) break;    }}int serch(int x){    int l=1,r=n-1,mid;    while(l<=r)    {        mid=l+r>>1;        if(tmp[mid].w>=x) l=mid+1;        else ans=mid,r=mid-1;    }return num-ans+1;}int main(){    freopen("warehouse.in","r",stdin);    freopen("warehouse.out","w",stdout);    int x,y,z,k;    n=read();m=read();q=read();    for(int i=1;i<=n;i++) fa[i]=i;    for(int i=1;i<=m;i++)    {        E[i].u=read();E[i].v=read();E[i].w=read();    }kruskal();    for(int i=1;i<=q;i++)    {        x=read();        if(x<=e[cnt].w){printf("1\n");continue;}        if(x>e[1].w){printf("%d\n",n);continue;}        k=serch(x);        printf("%d\n",k+1);    }    return 0;}