Incredible Crazily Progressing Company (ICPC) suffered a lot with the low speed of procedure. After investigation, they found that the bottleneck was at Absolutely Crowded Manufactory (ACM). In oder to accelerate the procedure, they bought a new machine for ACM. But a new problem comes, how to place the new machine into ACM?

ACM is a rectangular factor and can be divided into W * H cells. There are N retangular old machines in ACM and the new machine can not occupy any cell where there is old machines. The new machine needs M consecutive cells. Consecutive cells means some adjacent cells in a line. You are asked to calculate the number of ways to choose the place for the new machine.

Input

There are multiple test cases (no more than 50). The first line of each test case contains 4 integers W, H, N, M (1 ≤ W, H ≤ 107, 0 ≤ N ≤ 50000, 1 ≤ M ≤ 1000), indicating the width and the length of the room, the number of old machines and the size of the new machine. Then N lines follow, each of which contains 4 integers Xi1, Yi1, Xi2 and Yi2 (1 ≤ Xi1 ≤ Xi2 ≤ W, 1 ≤ Yi1 ≤ Yi2 ≤ H), indicating the coordinates of the i-th old machine. It is guarantees that no cell is occupied by two old machines.

Output

Output the number of ways to choose the cells to place the new machine in one line.

Sample Input

3 3 1 2
2 2 2 2
3 3 1 3
2 2 2 2
2 3 2 2
1 1 1 1
2 3 2 3

 

Sample Output

8
4
3
My Code 
  1 #include <cstdio>
  2 #include <algorithm>
  3 using namespace std;
  4 #define lson l,m,rt<<1
  5 #define rson m+1,r,rt<<1|1
  6 #define maxn 100100
  7 struct node
  8 {
  9     int len,c;
 10 }setree[maxn<<2];
 11 struct op
 12 {
 13     int l,r,cnt,h;
 14 }mes[maxn],mes1[maxn];
 15 int sorted[maxn],sorted1[maxn];
 16 long long ans;
 17 int w,h,n,m;
 18 bool cmp(struct op a,struct op b)
 19 {
 20     return a.h<b.h;
 21 }
 22 void build(int l,int r,int rt)
 23 {
 24     setree[rt].c=0;
 25     setree[rt].len=0;
 26     if(l==r)
 27     return;
 28     int m=(l+r)>>1;
 29     build(lson);
 30     build(rson);
 31 }
 32 int binsearch(int a[],int l,int r,int num)
 33 {
 34     int m=(l+r)>>1;
 35     if(a[m]==num)
 36     return m;
 37     if(num<a[m])
 38     return binsearch(a,l,m-1,num);
 39     return binsearch(a,m+1,r,num);
 40 }
 41 void pushup(int a[],int rt,int l,int r)
 42 {
 43     if(setree[rt].c>0)
 44     setree[rt].len=a[r]-a[l-1];
 45     else
 46     setree[rt].len=setree[rt<<1].len+setree[rt<<1|1].len;
 47     
 48 }
 49 void update(int a[],int l,int r,int rt,int L,int R,int c)
 50 {
 51     if(L<=l&&r<=R){
 52         setree[rt].c+=c;
 53         if(setree[rt].c>0)
 54         setree[rt].len=a[r]-a[l-1];
 55         else{
 56             if(l==r)
 57             setree[rt].len=0;
 58             else
 59             setree[rt].len=setree[rt<<1].len+setree[rt<<1|1].len;
 60         }
 61         return;
 62     }
 63     int m=(l+r)>>1;
 64     if(L<=m)
 65     update(a,lson,L,R,c);
 66     if(R>m)
 67     update(a,rson,L,R,c);
 68     pushup(a,rt,l,r);
 69 }
 70 void solve()
 71 {
 72     sort(mes,mes+2*n,cmp);
 73     sort(sorted,sorted+2*n);
 74     int k=1;
 75     long long s=0;
 76     for(int i=1;i<2*n;i++)
 77     if(sorted[i]!=sorted[i-1])
 78     sorted[k++]=sorted[i];
 79     build(1,k,1);
 80     for(int i=0;i<2*n;i++){
 81         int l=binsearch(sorted,0,k-1,mes[i].l);
 82         int r=binsearch(sorted,0,k-1,mes[i].r);
 83         update(sorted,1,k,1,l+1,r,mes[i].cnt);
 84         s+=(long long)(mes[i+1].h-mes[i].h)*setree[1].len;
 85     }
 86     ans+=(long long)w*h-s;
 87 }
 88 void solve1()
 89 {
 90     sort(mes1,mes1+2*n,cmp);
 91     sort(sorted1,sorted1+2*n);
 92     int k=1;
 93     long long s=0;
 94     for(int i=1;i<2*n;i++)
 95     if(sorted1[i]!=sorted1[i-1])
 96     sorted1[k++]=sorted1[i];
 97     build(1,k,1);
 98     for(int i=0;i<2*n;i++){
 99         int l=binsearch(sorted1,0,k-1,mes1[i].l);
100         int r=binsearch(sorted1,0,k-1,mes1[i].r);
101         update(sorted1,1,k,1,l+1,r,mes1[i].cnt);
102         s+=(long long)(mes1[i+1].h-mes1[i].h)*setree[1].len;
103     }
104     ans+=(long long)w*h-s;
105 }
106 int main()
107 {
108     while(~scanf("%d%d%d%d",&w,&h,&n,&m)){
109         ans=0;
110         for(int i=0;i<n;i++){
111             int a,b,c,d;
112             scanf("%d%d%d%d",&a,&b,&c,&d);
113             mes[2*i].l=a-m>0?a-m:0;mes[2*i].r=c;mes[2*i].cnt=1;mes[2*i].h=b-1;
114             mes[2*i+1].l=a-m>0?a-m:0;mes[2*i+1].r=c;mes[2*i+1].cnt=-1;mes[2*i+1].h=d;
115             sorted[2*i]=a-m>0?a-m:0;sorted[2*i+1]=c;
116             
117             mes1[2*i].l=a-1;mes1[2*i].r=c;mes1[2*i].cnt=1;mes1[2*i].h=b-m>0?b-m:0;
118             mes1[2*i+1].l=a-1;mes1[2*i+1].r=c;mes1[2*i+1].cnt=-1;mes1[2*i+1].h=d;
119             sorted1[2*i]=a-1;sorted1[2*i+1]=c;        
120         }
121         if(m>1){
122         mes[2*n].l=w-m+1>0?w-m+1:0;mes[2*n].r=w;mes[2*n].h=0;mes[2*n].cnt=1;
123         mes[2*n+1].l=w-m+1>0?w-m+1:0;mes[2*n+1].r=w;mes[2*n+1].h=h;mes[2*n+1].cnt=-1;
124         sorted[2*n]=w-m+1>0?w-m+1:0;sorted[2*n+1]=w;
125         
126         mes1[2*n].l=0;mes1[2*n].r=w;mes1[2*n].h=h-m+1>0?h-m+1:0;mes1[2*n].cnt=1;
127         mes1[2*n+1].l=0;mes1[2*n+1].r=w;mes1[2*n+1].h=h;mes1[2*n+1].cnt=-1;
128         sorted1[2*n]=0;sorted1[2*n+1]=w;
129         n++;
130         }
131         solve();
132         solve1();
133         if(m==1)
134         ans/=2;
135         printf("%lld\n",ans);
136     }
137 }