#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define l(a, b, i) for (ll i = a; i < b; i++)
#define rl(a, b, i) for (ll i = a; i >= b; i--)
#define vpair vector<pair<ll, ll>>
#define inf LLONG_MAX
#define ninf LLONG_MIN
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
ll N; cin >> N; vector<ll> vecA(N + 1, 0), vecB(N + 1, 0);
l(1, N + 1, i) cin >> vecA[i];
l(1, N + 1, i) cin >> vecB[i];
vector<ll> pre1(N + 1, 0); // cai prefix bthg
vector<ll> pre2(N + 1, 0); // prefix with wrap-around
// pre1[0] = pre1[1] = 0
// else, pre1[i] = pre1[i - 1] + vecB[i - 1]
l(2, N + 1, i) pre1[i] = pre1[i - 1] + vecB[i - 1];
// pre2[0] = 0
// pre2[1] = pre1[N] + vecB[N] // later need to minus away the pre1[idx we start at]
// else, pre2[i] = pre2[i - 1] + vecB[i - 1]
pre2[1] = pre1[N] + vecB[N];
l(2, N + 1, i) pre2[i] = pre2[i - 1] + vecB[i - 1];
vector<ll> need1(N + 1, 0); // vecA[i] - pre1[i]
vector<ll> need2(N + 1, 0); // vecA[i] - pre2[i] (lat nua cung can minus away the pre1[idx we start at])
l(1, N + 1, i) need1[i] = vecA[i] - pre1[i];
l(1, N + 1, i) need2[i] = vecA[i] - pre2[i];
vector<ll> maxneed1(N + 1, 0), maxneed2(N + 1, 0);
maxneed1[N] = need1[N];
rl(N - 1, 1, i) maxneed1[i] = max(maxneed1[i + 1], need1[i]);
maxneed2[1] = need2[1];
l(2, N + 1, i) maxneed2[i] = max(maxneed2[i - 1], need2[i]);
ll finalans = inf;
l(1, N + 1, i) {
ll need = max(maxneed1[i], maxneed2[i - 1]) + pre1[i];
finalans = min(finalans, need);
}
cout << finalans;
}
