#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define gcd __gcd
#define sz(v) (int) v.size()
#define pb push_back
#define pi pair<int,int>
#define all(v) (v).begin(), (v).end()
#define compact(v) (v).erase(unique(all(v)), (v).end())
#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define REV(i, a, b) for(int i = (a); i >= (b); i--)
#define dbg(x) "[" #x " = " << (x) << "]"
///#define int long long
using ll = long long;
using ld = long double;
using ull = unsigned long long;
template<typename T> bool ckmx(T& a, const T& b){if(a < b) return a = b, true; return false;}
template<typename T> bool ckmn(T& a, const T& b){if(a > b) return a = b, true; return false;}
const int N = 4e5+5;
const ll MOD = 1e9+7;
const ll INF = 1e18;
struct Item{
int l, r; ll c;
Item(int l = -1, int r = -1, ll c = 0): l(l), r(r), c(c) {}
};
ll n, m, val[2][N], dp[2][N], pref[N];
vector<Item> cross[2];
ll calc_dp(vector<Item>& cross, ll val[], ll dp[]){
FOR(i, 1, n + m - 1){
pref[i] = val[i];
pref[i] += pref[i - 1];
}
/*
+ dp[i] = how to fill all from 1 -> i with i must be fill by horizontal line
- choose all line that need to fill i
and the prefix from 1 -> i - 1 will be fill by vertical line (base case)
- for each j from 1 -> i - 1 we add horizontal line that need to fill i
and from j + 1 -> i - 1 will be fill by vertical line
*/
/*
to calculate answer
- we will iterate from 1 -> n and take min of
dp[i] + (vertical line will be fill from i+1 -> n)
- base case of answer is full of vertical with no horizontal
*/
ll answer = pref[n + m - 1];
FOR(i, 1, n + m - 1){
ll sum = 0;
vector<ll> lazy(n + m + 2);
FOR(j, 1, sz(cross) - 1){
if(cross[j].l <= i && i <= cross[j].r){
sum += cross[j].c;
lazy[1] += cross[j].c;
lazy[cross[j].l] -= cross[j].c;
}
}
// base case
dp[i] = sum + pref[i - 1];
FOR(j, 1, i - 1){
dp[i] = min(dp[i], dp[j] + lazy[j] +
pref[i - 1] - pref[j]);
lazy[j + 1] += lazy[j];
}
answer = min(answer, dp[i] + pref[n + m - 1] - pref[i]);
}
return answer;
}
struct SMT{
int trsz;
vector<ll> st, laz;
SMT(int n = 0): trsz(n), st((n << 2 | 1)), laz((n << 2 | 1)) {}
void apply(int id, ll val){
st[id] += val;
laz[id] += val;
}
void down(int id){
ll val = laz[id]; laz[id] = 0;
apply(id << 1, val);
apply(id << 1|1, val);
}
void update(int id, int l, int r, int u, int v, ll val){
if(l >= u && r <= v){
apply(id, val);
}
else{
int m = l+r>>1;
down(id);
if(u <= m) update(id << 1, l, m, u, v, val);
if(v > m) update(id << 1|1, m+1, r, u, v, val);
st[id] = min(st[id << 1], st[id << 1|1]);
}
}
ll get(int id, int l, int r, int u, int v){
if(l >= u && r <= v) return st[id];
down(id);
int m = l+r>>1;
ll answer = INF;
if(u <= m) ckmn(answer, get(id << 1, l, m, u, v));
if(v > m) ckmn(answer, get(id << 1|1, m+1, r, u, v));
return answer;
}
void update(int l, int r, ll val){
if(l > r) return;
update(1, 1, trsz, l, r, val);
}
ll query(int l, int r){
return get(1, 1, trsz, l, r);
}
void debugST(int id, int l, int r){
if(l == r){
// cout << l << " " << st[id] << "\n";
}
else{
int m = l+r>>1;
down(id);
debugST(id << 1, l, m);
debugST(id << 1|1, m+1, r);
}
}
};
ll opti_dp(vector<Item>& cross, ll val[]){
FOR(i, 1, n + m - 1){
pref[i] = val[i];
pref[i] += pref[i - 1];
}
vector<vector<ll>> add(n + m + 2, vector<ll>());
vector<vector<ll>> rem(n + m + 2, vector<ll>());
FOR(i, 1, sz(cross) - 1){
add[cross[i].l].push_back(i);
rem[cross[i].r + 1].push_back(i);
}
SMT dp(n + m - 1);
ll answer = pref[n + m - 1];
FOR(i, 1, n + m - 1){
for(int id : add[i]){
// we will need line id for 1 -> cross[id].r
//cout << "ADD: " << cross[id].r <<" " << cross[id].c << "\n";
dp.update(1, cross[id].r, cross[id].c);
}
for(int id : rem[i]){
// cross[id].l -> cross[id].r will now fill with vertical line
//cout << "DEL: " << cross[id].c << "\n";
dp.update(1, cross[id].l - 1, -cross[id].c);
}
dp.update(i, i, pref[i - 1]);
//dp.debugST(1, 1, n + m - 1);
ll ndp = dp.query(1, i);
answer = min(answer, ndp + pref[n + m - 1] - pref[i]);
dp.update(i, i, -dp.query(i, i)); // wtf how dumb ?
dp.update(i, i, ndp);
//cout << i <<" " << ndp << " " << dbg(pref[n + m - 1]) << dbg(pref[i]) << "\n";
dp.update(1, i - 1, val[i]);
}
//cout << answer << "\n";
//cout << "END DP\n\n";
return answer;
}
void solve()
{
cin >> m >> n;
cross[0].push_back(Item());
cross[1].push_back(Item());
FOR(i, 1, m + n - 1){
int l,r;
if(i <= n){
l = n - i + 1;
}
else l = i - n + 1;
if(i <= m){
r = n + i - 1;
}
else r = n + m - 1 - (i - m);
int cur_w; cin >> cur_w;
cross[(i & 1)].push_back(Item(l, r, cur_w));
}
FOR(i, 1, m + n - 1){
int cur_w; cin >> cur_w;
if((i & 1) == (n & 1)){
val[1][i] = cur_w;
}
else val[0][i] = cur_w;
}
ll answer = 0;
FOR(i, 0, 1){
answer += opti_dp(cross[i], val[i]);
//answer += calc_dp(cross[i], val[i], dp[i]);
}
cout << answer << "\n";
}
signed main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#define name "InvMOD"
if(fopen(name".INP", "r")){
freopen(name".INP","r",stdin);
freopen(name".OUT","w",stdout);
}
int t = 1; //cin >> t;
while(t--) solve();
return 0;
}
/* Test
Input:
4 3
5 3 4 3 5 3
5 2 5 2 4 2
Output: 19
Input:
3 3
4 1 2 2 3
4 1 2 1 4
Output: 6
*/
