/*题目大意：输入一序列n个数字，然后输入m个询问，每个询问包含左边区间和右边区间，还有a和b，问你这个区间内有几个数大于等于a且小于等于b

 做法：树状数组，先求出这个区间内有几个数小于a，然后求这个区间内有几个数小于等于b，拿后者减去前者就是答案了*/

 #include<stdio.h>

 #include<string.h>

 #include<algorithm>

 using namespace std;

 const int maxn=;

 int c[maxn],ansl[maxn],ansr[maxn];

 struct pp

 {

     int num;

     int id;

 }p[maxn];

 struct p2p

 {

     int hi;

     int id;

     int l,r;

     int low;

 }p1[maxn];

 int cmp(pp a,pp b)

 {

         return a.num<b.num;

 }

 int cmp1(p2p a,p2p b)

 {

     return a.low<b.low;

 }

 int cmp2(p2p a,p2p b)

 {

     return a.hi<b.hi;

 }

 int n,m;

 int lowbit(int x)

 {

     return x&(-x);

 }

 void add(int x,int d)

 {

     while(x<=n)

     {

         c[x]+=d;

         x=x+lowbit(x);

     }

 }

 int sum(int x)

 {

     int ans=;

     while(x>)

     {

         ans+=c[x];

         x=x-lowbit(x);

     }

   return ans;

 }

 int main()

 {

     int i,j,k;

     int t;

     int cas=;

     scanf("%d",&t);

     while(t--)

     {

         memset(c,,sizeof(c));

         memset(ansl,,sizeof(ansl));

         scanf("%d%d",&n,&m);

         for(i=;i<=n;i++)

         {

             scanf("%d",&p[i].num);

             p[i].id=i;

         }

         for(i=;i<=m;i++)

         {

             scanf("%d%d%d%d",&p1[i].l,&p1[i].r,&p1[i].low,&p1[i].hi);

             p1[i].id=i;

         }

         sort(p+,p+n+,cmp);

         sort(p1+,p1+m+,cmp1);

         i=;

         j=;

        while(j<=m)

       {

           while(i<=n)

           {

             if(p[i].num>=p1[j].low)

             break;

             add(p[i].id,);

             i++;

           }

          while(j<=m)

          {

           if(i<=n&&p[i].num<p1[j].low)

           break;

           ansl[p1[j].id]=sum(p1[j].r)-sum(p1[j].l-);

           j++;

          }

       }

       i=;

       j=;

       memset(c,,sizeof(c));

       memset(ansr,,sizeof(ansr));

       sort(p1+,p1+m+,cmp2);

       while(j<=m)

       {

           while(i<=n)

           {

             if(p[i].num>p1[j].hi)

             break;

             add(p[i].id,);

             i++;

           }

          while(j<=m)

          {

           if(i<=n&&p[i].num<=p1[j].hi)

           break;

           ansr[p1[j].id]=sum(p1[j].r)-sum(p1[j].l-);

           j++;

          }

       }

       printf("Case #%d:\n",++cas);

       for(i=;i<=m;i++)

           if(ansr[i]-ansl[i]>)

           printf("%d\n",ansr[i]-ansl[i]);

           else printf("0\n");

     }

     return ;

 }