#include <bits/stdc++.h>
using namespace std;
int n,m,d;
map<pair<int,int>,vector<int>> onpoz;
int poz[500005],cntactive;
bool active[500005];
pair<int,int> special[500005];
int goup[10005],goleft[10005];
signed main()
{
ios_base::sync_with_stdio(0);cin.tie(0);
cin>>n>>m>>d;
for(int i=0;i<n;i++)
{
int p,q;
cin>>p>>q;
onpoz[{p,q}].push_back(i);
onpoz[{p,q+d}].push_back(i);
onpoz[{p+d,q}].push_back(i);
onpoz[{p+d,q+d}].push_back(i);
}
for(int i=0;i<m;i++)
{
int p,q;
cin>>p>>q;
special[i] = {p,q};
}
int rez = d*d;
for(int jos=0;jos<2*d;jos++)
{
for(int i=0;i<d;i++)
{
for(int x:onpoz[{jos,i}])
{
if(!active[x])
{
cntactive++;
active[x] = 1;
}
poz[x] = jos;
}
}
if(cntactive == n)
{
int minlin=jos;
for(int x=0;x<n;x++)
minlin = min(minlin, poz[x]);
goup[jos] = minlin;
}
else
goup[jos] = -1;
}
cntactive=0;
for(int x=0;x<n;x++)
active[x]=0;
for(int jos=0;jos<2*d;jos++)
{
for(int i=0;i<d;i++)
{
for(int x:onpoz[{i,jos}])
{
if(!active[x])
{
cntactive++;
active[x] = 1;
}
poz[x] = jos;
}
}
if(cntactive == n)
{
int minlin=jos;
for(int x=0;x<n;x++)
minlin = min(minlin, poz[x]);
goleft[jos] = minlin;
}
else
goleft[jos] = -1;
}
for(int jos=d;jos<2*d;jos++)
{
for(int sus=jos-d+1;sus<=goup[jos];sus++)
{
vector<int> pozs;
for(int x=0;x<m;x++)
{
if((sus <= special[x].first && special[x].first <= jos) || (sus <= special[x].first + d && special[x].first + d <= jos))
{
}
else
{
pozs.push_back(special[x].second);
}
}
int mnm = d;
if(pozs.empty())
{
}
else
{
sort(pozs.begin(),pozs.end());
for(int i=0;i+1<pozs.size();i++)
{
mnm = min(mnm, pozs[i]+d - min(goleft[pozs[i]+d], pozs[i+1]) + 1);
}
mnm = min(mnm, pozs.back()+d - min(goleft[pozs.back()+d], pozs[0]+d) + 1);
}
rez = min(rez, (jos-sus+1) * mnm);
}
}
cout<<rez;
return 0;
}
/*
2 1 5
1 4
2 2
0 0
patratele se repeta din d in d
obs: o sa existe tot timpu o solutie care are fiecare latura <d
vrem sa facem O(d) verificari de tipu: putem incadra punctele intr-un dreptunghi de dimensiunea Lx pe Ly
*/
