1 //minamoto
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #define int long long
 6 using namespace std;
 7 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
 8 char buf[1<<21],*p1=buf,*p2=buf;
 9 inline int read(){
10     #define num ch-'0'
11     char ch;bool flag=0;int res;
12     while(!isdigit(ch=getc()))
13     (ch=='-')&&(flag=true);
14     for(res=num;isdigit(ch=getc());res=res*10+num);
15     (flag)&&(res=-res);
16     #undef num
17     return res;
18 }
19 const int N=100005,mod=2147483648;
20 int n,m,k,C[N][15],tt=0,ans,tmp[N],c[N],col,tot[N],cnt[N],r[N],h[N];
21 struct node{int x,y;}a[N];
22 inline bool cmp(const node &a,const node &b)
23 {return a.x!=b.x?a.x<b.x:a.y<b.y;}
24 inline bool cmp2(const node &a,const node &b)
25 {return a.y!=b.y?a.y<b.y:a.x<b.x;}
26 inline void add(int x,int y){
27     for(;x<=col;x+=x&-x) (c[x]+=y)%=mod;
28 }
29 inline int query(int x){
30     int res=0;
31     for(;x;x-=x&-x) (res+=c[x])%=mod;
32     return res;
33 }
34 signed main(){
35     //freopen("testdata.in","r",stdin);
36     read();read();n=read();
37     for(int i=1;i<=n;++i) a[i].x=read(),a[i].y=read();
38     k=read();
39     for(int i=0;i<=n;++i) C[i][0]=1;
40     for(int i=1;i<=n;++i)
41     for(int j=1;j<=min(i,k);++j)
42     C[i][j]=C[i-1][j]+C[i-1][j-1];
43     sort(a+1,a+1+n,cmp2);
44     for(int i=1;i<=n;++i)
45     tmp[i]=(i==1||a[i].y!=a[i-1].y)?++tt:tt;
46     for(int i=1;i<=n;++i) cnt[a[i].y=tmp[i]]++;col=a[n].y;
47     sort(a+1,a+1+n,cmp);
48     for(int i=1;i<=n;++i)
49     tmp[i]=(i==1||a[i].x!=a[i-1].x)?++tt:tt;
50     for(int i=1;i<=n;++i) tot[a[i].x=tmp[i]]++;
51     for(int i=1;i<=n;++i){
52         if(i==1||a[i].x!=a[i-1].x) tt=0;
53         int dy=a[i].y,v=(++h[dy])>=k&&cnt[dy]-h[dy]>=k?
54         1ll*C[h[dy]][k]*C[cnt[dy]-h[dy]][k]%mod:0;++tt;
55         add(dy,v-r[dy]),r[dy]=v;
56         if(i==n||a[i].x!=a[i+1].x||a[i+1].y-a[i].y<=1
57         ||tt<k||tot[a[i].x]-tt<k) continue;
58         (ans+=1ll*C[tt][k]*C[tot[a[i].x]-tt][k]%mod
59         *(query(a[i+1].y-1)-query(a[i].y)))%=mod;
60     }
61     printf("%d\n",(ans>=0?ans:ans+mod)%mod);
62     return 0;
63 }