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#include <bits/stdc++.h>
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using namespace std;
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const int N = 100010, M = 2000010; // 因为要建新图，两倍的边
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int n, m; // 点数、边数
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int dfn[N], low[N], ts;
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int stk[N], top, in_stk[N];
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int id[N], scc_cnt, sz[N];
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int f[N];
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int h[N], hs[N], e[M], ne[M], idx; // h: 原图；hs: 新图
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void add(int h[], int a, int b) {
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e[idx] = b, ne[idx] = h[a], h[a] = idx++;
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}
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int p[N]; //原图的点权
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int q[N]; //新图的点权
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// tarjan求强连通分量
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void tarjan(int u) {
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dfn[u] = low[u] = ++ts;
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stk[++top] = u;
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in_stk[u] = true;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (!dfn[j]) {
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tarjan(j);
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low[u] = min(low[u], low[j]);
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} else if (in_stk[j])
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low[u] = min(low[u], dfn[j]);
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}
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if (dfn[u] == low[u]) {
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++scc_cnt;
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int x;
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do {
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x = stk[top--];
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in_stk[x] = false;
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id[x] = scc_cnt;
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sz[scc_cnt]++;
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// scc_cnt:强连通块的编号
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q[scc_cnt] += p[x]; //叠加点权，生成连通块的总点权
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} while (x != u);
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}
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}
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int main() {
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memset(h, -1, sizeof h); //原图
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memset(hs, -1, sizeof hs); //新图
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scanf("%d %d", &n, &m);
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//读入点权
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for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
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for (int i = 1; i <= m; i++) {
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int a, b;
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scanf("%d %d", &a, &b);
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add(h, a, b);
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}
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//利用强连通分量，缩点，生成DAG
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i);
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// (2) 缩点，建图
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for (int u = 1; u <= n; u++)
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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int a = id[u], b = id[j];
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if (a != b) //去重边
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add(hs, a, b); //加入到新图
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}
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// (3) 根据拓扑序遍历DAG，从scc_cnt向前遍历自然满足拓扑序
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for (int u = scc_cnt; u; u--) {
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// base case 递推起点
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if (!f[u]) f[u] = q[u];
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for (int i = hs[u]; ~i; i = ne[i]) {
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int j = e[i]; // 边(i, j)
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if (f[j] < f[u] + q[j]) f[j] = f[u] + q[j];
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}
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}
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// (4) 求解答案
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int res = 0;
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for (int i = 1; i <= scc_cnt; i++) res = max(res, f[i]);
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//输出
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printf("%d\n", res);
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return 0;
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}
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