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對於點x，從S向x連容量為va的邊，從x向T連容量為vb的邊。

對於邊(u, v)，從S向u和v分別連兩條容量為ea / 2的邊，從u和v向T分別連兩條容量為eb / 2的邊，然後u和v中間互相連兩條容量為ea / 2 + eb / 2 + ec的邊。

答案是 最大貢獻 減去 最小割。

#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 10005, maxm = 40005, maxq = 50000, inf = 0x3f3f3f3f;

int n, m, head[maxn], cur[maxn], cnt, bg, ed, depth[maxn], q[maxq];

struct _edge {
	int v, w, next;
} g[4 * maxn + 10 * maxm];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v, int w) {
	g[cnt] = (_edge) {v, w, head[u]};
	head[u] = cnt++;
}

bool bfs() {
	for(int i = bg; i <= ed; i++) depth[i] = -1;
	int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1;
	while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) {
		depth[g[i].v] = depth[u] + 1;
		if(g[i].v == ed) return 1;
		q[t++] = g[i].v;
	}
	return 0;
}

int dfs(int x, int flow) {
	if(x == ed) return flow;
	int left = flow;
	for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) {
		int tmp = dfs(g[i].v, min(left, g[i].w));
		left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp;
		if(g[i].w) cur[x] = i;
		if(!left) return flow;
	}
	if(left == flow) depth[x] = -1;
	return flow - left;
}

int main() {
	n = iread(); m = iread(); bg = 0; ed = n + 1;
	for(int i = bg; i <= ed; i++) head[i] = -1;

	LL ans = 0;
	add(bg, 1, inf); add(1, bg, 0); add(n, ed, inf); add(ed, n, 0);
	for(int i = 2; i < n; i++) {
		int va = iread(); va <<= 1;
		add(bg, i, va); add(i, bg, 0);
		ans += va;
	}
	for(int i = 2; i < n; i++) {
		int vb = iread(); vb <<= 1;
		add(i, ed, vb); add(ed, i, 0);
		ans += vb;
	}
	for(int i = 1; i <= m; i++) {
		int x = iread(), y = iread(), ea = iread(), eb = iread(), ec = iread();
		add(bg, x, ea); add(x, bg, 0);
		add(bg, y, ea); add(y, bg, 0);
		add(x, ed, eb); add(ed, x, 0);
		add(y, ed, eb); add(ed, y, 0);
		add(x, y, ea + eb + (ec << 1)); add(y, x, ea + eb + (ec << 1));
		ans += ea + eb << 1;
	}

	while(bfs()) {
		for(int i = bg; i <= ed; i++) cur[i] = head[i];
		ans -= dfs(bg, inf);
	}

	printf("%lld\n", ans >> 1);
	return 0;
}