#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void prop(vector<array<ll, 2>> &ls, ll at, ll atl, ll atr){
if(!ls[at][0]) return;
ls[at][1] += ls[at][0] * (atr - atl + 1);
if(atl != atr)
ls[2 * at][0] += ls[at][0], ls[2 * at + 1][0] += ls[at][0];
ls[at][0] = 0;
}
void up(vector<array<ll, 2>> &ls, ll at, ll atl, ll atr, ll lq, ll rq, ll vl){
prop(ls, at, atl, atr);
if(rq < atl || atr < lq) return;
if(lq <= atl && atr <= rq){
ls[at][0] = vl;
prop(ls, at, atl, atr);
return;
}
ll m = (atl + atr) / 2;
up(ls, 2 * at, atl, m, lq, rq, vl), up(ls, 2 * at + 1, m + 1, atr, lq, rq, vl);
ls[at][1] = min(ls[2 * at][1], ls[2 * at + 1][1]);
}
ll qr(vector<array<ll, 2>> &ls, ll at, ll atl, ll atr, ll lq, ll rq){
prop(ls, at, atl, atr);
if(rq < atl || atr < lq) return 1e16;
if(lq <= atl && atr <= rq) return ls[at][1];
ll m = (atl + atr) / 2, r1 = qr(ls, 2 * at, atl, m, lq, rq);
return min(r1, qr(ls, 2 * at + 1, m + 1, atr, lq, rq));
}
ll sol(vector<ll> a, vector<array<ll, 3>> b){
ll sz = a.size() + 1;
vector<ll> dp(sz); // seg com lazy
vector<vector<array<ll, 2>>> sl(sz + 1); // 1 indexado e um espaco a mais pra não dar segfal
vector<array<ll, 2>> ls(4 * sz, {0, 0});
for(ll i = 0; i < sz - 2; i++){ // pro sweep line
sl[b[i][0]].push_back({-1, i});
sl[b[i][1] + 1].push_back({1, i});
}
ll s = 0;
for(ll i = 1; 1; i++){
for(auto [j, k] : sl[i]){
up(ls, 1, 0, sz, b[k][0], b[k][1], j * b[k][2]);
s -= j * b[k][2];
}
if(i == sz - 1) break;
ll vl = qr(ls, 1, 0, sz, 0, i - 1);
dp[i] = s + vl;
up(ls, 1, 0, sz, 0, i - 1, a[i]); // aviso pro proximo que posso não ligar nenhum
up(ls, 1, 0, sz, i, i, dp[i]);
}
ll re = 1e16;
for(ll i = 0; i < sz - 1; i++)
re = min(re, qr(ls, 1, 0, sz, i, i));
return re;
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
ll n, m; cin >> n >> m;
vector<array<ll, 3>> b, p;
vector<ll> b2 = {-1}, p2 = b2;
for(ll i = 1, vl, l = m, r = m, dl = -1, dr = 1; i <= n + m - 1; i++, l += dl, r += dr){
cin >> vl;
if(i % 2) b.push_back({(l + 1) / 2, (r + 1) / 2, vl});
else p.push_back({(l + 1) / 2, (r + 1) / 2, vl});
if(l == 1) dl = 1;
if(r == n + m - 1) dr = -1;
}
for(ll i = 1, vl; i <= n + m - 1; i++){
cin >> vl;
if(i % 2 == m % 2) b2.push_back(vl);
else p2.push_back(vl);
}
cout << sol(b2, b) + sol(p2, p) << "\n"; // esboco, termino amanha ta muito errado
}
