To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn't tan at all........

The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPF_i (1 ≤ SPF_i ≤ 1,000). Lotion bottle i can cover cover_i cows with lotion. A cow may lotion from only one bottle.

What is the maximum number of cows that can protect themselves while tanning given the available lotions?

* Line 1: Two space-separated integers: C and L
* Lines 2..C+1: Line i describes cow i's lotion requires with two integers: minSPF_i and maxSPF_i 
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPF_i and cover_i

A single line with an integer that is the maximum number of cows that can be protected while tanning

 #include <iostream>

 #include <cstring>

 #include <queue>

 #include <algorithm>

 using namespace std;

 struct node

 {

     int a,b;

 }cow[],s[];

 bool cmp(node x,node y)

 {

     return x.a<=y.a;

 }

 int main()

 {

     int c,l;

     int i,j;

     priority_queue<int,vector<int>,greater<int> > Q;

     freopen("in.txt","r",stdin);

     scanf("%d%d",&c,&l);

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

         scanf("%d%d",&cow[i].a,&cow[i].b);

     for(j=;j<l;j++)

         scanf("%d%d",&s[j].a,&s[j].b);

     sort(cow,cow+c,cmp);

     sort(s,s+l,cmp);

     j=;

     int sum=;

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

     {

         while(j<c&&cow[j].a<=s[i].a)

         {

             Q.push(cow[j].b);

             j++;

         }

         while(!Q.empty()&&s[i].b)

         {

             int t=Q.top();

             Q.pop();

             if(s[i].a<=t)

             {

                 sum++;

                 s[i].b--;

             }

         }

     }

     printf("%d\n",sum);

 }