/* Author : Mychecksdead */
#include<bits/stdc++.h>
// #pragma GCC optimize("O3,unroll-loops")
// #pragma GCC target("avx2,bmi,bmi2")
using namespace std;
#define ll long long int
#define MOD (1000000000+7)
#define MOD1 (998244353)
#define pb push_back
#define all(x) x.begin(), x.end()
#define en cout << '\n'
#define ff first
#define ss second
#define pii pair<int,int>
#define vi vector<int>
#define cerr if(false) cerr
const int N = 1e6+100, M = 1e5+10, K = 52, MX = 30;
int n, m, D, mx[N], L[N], R[N];
array<int, 2> a[N], b[N], c[N];
vector<int> cnt(N), V(N, -1);
void add(int y, int idx){
cnt[y]++;
V[y] = idx; // only the last one is important
}
void del(int y){
cnt[y]--;
}
void process(){
int cur = 0, last = -1, mn = MOD;
for(int i = 0; i < D; ++i){
cerr << cnt[i] << ' ';
if(cnt[i]){
if(mn == MOD) mn = i;
if(last != -1){
cur = max(cur, i - last);
}
last = i;
}
}
cerr << '\n';
cur = max(cur, mn + D - last);
mx[0] = cur - 1;
cerr << mn << ' ' << last<< ' ' << mx[0] << " f\n";
// mx[i] = pt=1...i is popped, what's the answer
vector<pii> Q;
for(int i = 0; i < D; ++i) if(V[i] != -1) Q.pb({i - D, V[i]});
for(int i = 0; i < D; ++i){
if(V[i] != -1){
while(Q.size() && Q.back().ss < V[i]) Q.pop_back();
if(Q.size()){
L[i] = Q.back().ff;
}else assert(false);
Q.pb({i, V[i]});
}
}
Q.clear();
for(int i = D-1; i >= 0; --i) if(V[i] != -1) Q.pb({i + D, V[i]});
for(int i = D-1; i >= 0; --i){
if(V[i] != -1){
while(Q.size() && Q.back().ss < V[i]) Q.pop_back();
if(Q.size()){
R[i] = Q.back().ff;
}else assert(false);
Q.pb({i, V[i]});
}
}
for(int i = 1; i <= max(D, m); ++i) mx[i] = -1;
for(int i = 0; i < D; ++i) cerr << V[i] << ' '<< L[i] << ' ' << R[i] << '\n';
for(int i = 0; i < D; ++i) if(V[i] >= 0 && V[i] < MOD) mx[V[i]] = max(mx[V[i]], R[i] - L[i] - 1);
for(int i = 1; i <= max(D, m); ++i) mx[i] = max(mx[i], mx[i - 1]);
}
void solve(){
cin >> n >> m >> D;
for(int i = 1; i <= n; ++i){
cin >> a[i][0] >> a[i][1];
c[i] = {a[i][1], a[i][0]};
}
for(int i = 1; i <= m; ++i){
cin >> b[i][0] >> b[i][1];
}
sort(a+1, a+1+n);
sort(c+1, c+1+n);
sort(b+1, b+1+m);
vi A;
for(int i = 1; i <= n; ++i){
if(i == 1 || c[i - 1][0] != c[i][0]){
A.pb(c[i][0]);
}
}
// add(c[1][0] + D, 0);
// A.pb(c[1][0] + D);
int R = a[n][0];
int ptr = 1;
int ans = D * D;
int ptr2 = 1;
for(int l = 0; l < D; ++l){
while(ptr <= n && l > a[ptr][0]){
R = a[ptr][0] + D;
++ptr;
}
int r = R; // minimum right border basically
cerr << l << ' ' << r << " :\n";
while(ptr2 <= m && l > b[ptr2][0]){
b[ptr2 + m] = {b[ptr2][0] + D, b[ptr2][1]};
++ptr2;
}
// we will try all type B's...
for(int i = ptr2, pt = 1; i < ptr2 + m; ++i) if(b[i][0] > r) add(b[i][1], pt++); // currently all B's are horizontal
for(int x: A) add(x, MOD);
process();
ans = min(ans, (D - mx[0]) * (r - l + 1));
cerr << D-mx[0] << ' ' << (r - l + 1) << " hmm\n";
cerr << mx[0] << " hell\n";
// this is the TLE area...
for(int i = ptr2, pt = 1; i < ptr2 + m; ++i){ // we pop one by one
if(b[i][0] > r)
ans = min(ans, (D - mx[pt++]) * (max(r, b[i][0]) - l + 1));
}
V.clear();
V.resize(D, -1);
for(int i = ptr2; i < ptr2 + m; ++i) if(b[i][0] > r) del(b[i][1]); // currently all B's are horizontal
}
cout << ans;
}
int main(){
cin.tie(0); ios::sync_with_stdio(0);
int tt = 1, aa;
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
while(tt--){
solve();
en;
}
cerr<<"time taken : "<<(float)clock()/CLOCKS_PER_SEC<<" seconds\n";
return 0;
}
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