#include<string.h>
#include<stdio.h>
#include<iostream>
#include <queue>
using namespace std;
const int N = 3005;
struct Edge
{
	int cost, next, to;
}e[N<<1];
int head[N<<1], cnt = 0;
int s[N];
void add(int from, int to, int cost)
{
	e[++cnt].to = to;
	e[cnt].cost = cost;
	e[cnt].next = head[from];
	head[from] = cnt;
}
int dp[N][N];
int n, m;
//dp[i][j] 以i为根，容量为j的最大值 总量还是以1为根的m次
void dfs(int x,int fx,int tot)
{
	for (int i = head[x]; i; i = e[i].next)
	{
		int v = e[i].to;
		if (v == fx)continue;
		for (int k = 0; k < tot; k++)
		{
			dp[v][k] = dp[x][k] - e[i].cost + s[v];
		}
		if (!s[v]) tot++;
			dfs(v, x, tot - 1);
		for (int k = 1; k <= tot; k++)
			dp[x][k] = max(dp[v][k - 1], dp[x][k]);
	}
}
int main()
{
	scanf("%d%d", &n,&m);
	for (int i = 1,l,k,w; i <= n-m; i++)
	{
		scanf("%d", &k);
		for (int j = 1; j <= k; j++)
		{
			scanf("%d%d", &l, &w);
			add(i, l, w);
		}
	}
	for (int i = 1; i <= m; i++)
		scanf("%d", s + i);
	int ans = 0;
	dfs(1, 0, m);
	for (int i = m; i >=1; i--)
	{
		if (dp[1][i] > 0)
		{
			ans = i;
			break;
		}
	}
	printf("%d",ans);
	return 0;
}