#include <iostream>
#include <bits/stdc++.h>
using namespace std;
// sample input:-
// 4
// 1 3 2 5
// 3 5 3 6
// 2
// 1 3
// 4 6
// answer should be 7
int main() {
// your code goes here
int n;
cin>> n;
vector<int> st(n,0), end(n,0);
// track max;
int mx=0;
for(int i=0; i<n; i++){
cin>>st[i];
}
for(int i=0; i<n; i++){
cin>>end[i];
mx = max(mx, end[i]);
}
int k;
cin>> k;
int qs[2], qe[2];
cin>>qs[0];
cin>>qs[1];
cin>>qe[0];
cin>>qe[1];
// <<<<<<<<<<<--------------------------->>>>>>>>>>>>
// solution
// 1 indexed prefix sum that's why (mx+1)
vector<int> pre (mx+1, 0);
// update pre with 1 or -1
for(int i=0; i<n; i++){
// from st
pre[st[i]] += 1;
// from end. only skip if it's the last element
// because last +1 doesn't exist to put -1
if(end[i]!=mx) pre[end[i]+1] += -1;
}
// calculate pre
for(int i=1; i<=mx; i++){
pre[i]+=pre[i-1];
}
int ans=0;
// if(pre[i]>=k) ans+=pre[i];
// find each hour which is in range of both the queries.
// queries might be overlapping. so
vector<int> arr(mx+1, 0);
for(int i=qs[0]; i<=qe[0]; i++){
if(pre[i]>=k) arr[i]++;
}
for(int i=qs[1]; i<=qe[1]; i++){
if(pre[i]>=k) arr[i]++;
}
// calculate for each one in the range of queries
// and add that hour's pre value in ans
for(int i=0; i<mx+1; i++){
if(arr[i]!=0) ans+=pre[i];
}
// for(auto x: pre) cout<<x<<endl;
cout<< ans;
return 0;
}
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