#include <bits/stdc++.h>

using namespace std;
const int N = 100010, M = 2000010; // 因为要建新图，两倍的边
int n, m;                          // 点数、边数
int dfn[N], low[N], ts;
int stk[N], top, in_stk[N];
int id[N], scc_cnt, sz[N];
int f[N];
int h[N], hs[N], e[M], ne[M], idx; // h: 原图；hs: 新图
void add(int h[], int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
int p[N]; //原图的点权
int q[N]; //新图的点权

// tarjan求强连通分量
void tarjan(int u) {
    dfn[u] = low[u] = ++ts;
    stk[++top] = u;
    in_stk[u] = true;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (!dfn[j]) {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        } else if (in_stk[j])
            low[u] = min(low[u], dfn[j]);
    }
    if (dfn[u] == low[u]) {
        ++scc_cnt;
        int x;
        do {
            x = stk[top--];
            in_stk[x] = false;
            id[x] = scc_cnt;
            sz[scc_cnt]++;
            // scc_cnt:强连通块的编号
            q[scc_cnt] += p[x]; //叠加点权，生成连通块的总点权
        } while (x != u);
    }
}

int main() {
    memset(h, -1, sizeof h);   //原图
    memset(hs, -1, sizeof hs); //新图

    scanf("%d %d", &n, &m);
    //读入点权
    for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
    for (int i = 1; i <= m; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        add(h, a, b);
    }
    //利用强连通分量，缩点，生成DAG
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);

    // (2) 缩点，建图
    for (int u = 1; u <= n; u++)
        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            int a = id[u], b = id[j];
            if (a != b)        //去重边
                add(hs, a, b); //加入到新图
        }

    // (3) 根据拓扑序遍历DAG，从scc_cnt向前遍历自然满足拓扑序
    for (int u = scc_cnt; u; u--) {
        // base case 递推起点
        if (!f[u]) f[u] = q[u];

        for (int i = hs[u]; ~i; i = ne[i]) {
            int j = e[i]; // 边(i, j)
            if (f[j] < f[u] + q[j]) f[j] = f[u] + q[j];
        }
    }

    // (4) 求解答案
    int res = 0;
    for (int i = 1; i <= scc_cnt; i++) res = max(res, f[i]);

    //输出
    printf("%d\n", res);
    return 0;
}