/*I题 1 or 2
对于边x,y建立两条边(x, y + n, 1), (y, x + n, 1),在s和i之间建边(s, i, d[i]),在i+n和t之间建边(i + n, t, d[i]),
然后从s到t跑一个最大流,判断是不是等于sum(d[1~n]),如果相等输出Yes,否则输出No
*/
#include<bits/stdc++.h>
using namespace std;
const int inf = 1 << 30, N = 5e4 + 7, M = 3e5 + 7;
int head[N], ver[M], edge[M], nex[M], d[N], cur[N];
int n, m, s, t, tot, maxflow;
queue<int> q;
void init() {
maxflow = 0;
tot = 1;
memset(head, 0, sizeof(head));
}
void add(int x, int y, int z) {
ver[++tot] = y, edge[tot] = z, nex[tot] = head[x], head[x] = tot;
ver[++tot] = x, edge[tot] = 0, nex[tot] = head[y], head[y] = tot;
}
bool bfs() {
memset(d, 0, sizeof(d));
while (q.size()) q.pop();
q.push(s); d[s] = 1; cur[s] = head[s];
while (q.size()) {
int x = q.front(); q.pop();
for (int i = head[x]; i; i = nex[i]) {
int y = ver[i];
if (edge[i] && !d[y]) {
q.push(y);
cur[y] = head[y];
d[y] = d[x] + 1;
if (y == t) return true;
}
}
}
return false;
}
int dinic(int x, int flow) {
if (x == t) return flow;
int rest = flow, k, i;
for (i = cur[x]; i && rest; i = nex[i]) {
int y = ver[i];
if (edge[i] && d[y] == d[x] + 1) {
k = dinic(y, min(rest, edge[i]));
if (!k) d[y] = 0;
edge[i] -= k;
edge[i ^ 1] += k;
rest -= k;
}
}
cur[x] = i;
return flow - rest;
}
int main() {
while (scanf("%d%d", &n, &m) != EOF) {
int p, sum = 0, x, y; s = 2 * n + 1; t = 2 * n + 2;
init();
for (int i = 1; i <= n; i++) {
scanf("%d", &p);
add(s, i, p);
add(i + n, t, p);
sum += p;
}
for (int i = 1; i <= m; i++) {
scanf("%d%d", &x, &y);
add(x, y + n, 1);
add(y, x + n, 1);
}
int flow;
while (bfs()) {
while ((flow = dinic(s, inf)))
maxflow += flow;
}
if (maxflow == sum)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}