#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int N = 1010, M = 5010;
int n, m;
int f[N], cnt[N];
double dist[N];
bool st[N];

// 邻接表
int idx, h[N], e[M], w[M], ne[M];
void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

bool check(double mid) {
    queue<int> q;
    memset(cnt, 0, sizeof cnt);
    memset(dist, 0x3f, sizeof dist);
    memset(st, false, sizeof st);
    for (int i = 1; i <= n; i++) {
        q.push(i);
        st[i] = 1;
    }

    while (q.size()) {
        int u = q.front();
        q.pop();
        st[u] = 0;
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            // 最短路
            if (dist[v] > dist[u] + w[i] * mid - f[u]) {
                dist[v] = dist[u] + w[i] * mid - f[u];
                // 判负环
                cnt[v] = cnt[u] + 1;
                if (cnt[v] >= n) return 1;
                if (!st[v]) {
                    q.push(v);
                    st[v] = 1;
                }
            }
        }
    }
    return 0;
}
int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &f[i]); // 每个点都有一个权值f[i]

    // 初始化邻接表
    memset(h, -1, sizeof h);
    int a, b, c;
    while (m--) {
        scanf("%d %d %d", &a, &b, &c);
        add(a, b, c);
    }

    // 浮点数二分
    double l = 0, r = INF;
    while (r - l > eps) {
        double mid = (l + r) / 2;
        if (check(mid))
            l = mid; // 存在负环时，mid再大一点,最终取得01分数规则的最大值
        else
            r = mid; // 不存在负环时，mid再小一点
    }
    printf("%.2lf\n", l);
    return 0;
}