#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
/**
 * https://www.acwing.com/problem/content/description/4398/
 * 思路:
 * 首先,不用把矩阵构造出来,子矩阵的面积和就是2个数组区间和的乘积
 * 预处理出每个区间长度下,区间和的最小值
 * 枚举区间每个长度,算出结果
 */
const int N = 2005;
int n,m;
int s1[N],s2[N];
int a[N],b[N];
int main(){
 cin>>n>>m;
 for(int i = 1; i <=n; i++) {
 int x;
 cin>>x;
 s1[i] = s1[i-1]+x;
 }
 for(int i = 1; i <=m; i++) {
 int x;
 cin>>x;
 s2[i] = s2[i-1]+x;
 }
 int limit;
 cin>>limit;
 //枚举每个区间长度
 memset(a,0x3f,sizeof a);
 memset(b,0x3f,sizeof b);
 for(int i = 1; i <=n; i++) {
 for(int j = 1; j+i-1 <=n; j++) {
 a[i] = min(s1[j+i-1]-s1[j-1],a[i]);
 }
 }
 for(int i = 1; i <=m; i++) {
 for(int j = 1; j+i-1 <=m; j++) {
 b[i] = min(s2[j+i-1]-s2[j-1],b[i]);
 }
 }
 int ans = 0;
 for(int i = n; i; i--) {
 for(int j = m; j ; j--) {
 //注意这里相乘会报int
 if ((LL)a[i]*b[j]<=limit){
 ans = max(ans,i*j);
 }
 }
 }
 cout<<ans;
}