#include<bits/stdc++.h>

using namespace std;

#define x first
#define y second
#define pb push_back
#define eb emplace_back
#define rep(i,a,b) for(auto i = (a); i != (b); ++i)
#define REP(i,n) rep(i,0,n)
#define all(v) (v).begin(), (v).end()

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int maxn = 1e6;

int N, M, C[maxn], D[maxn] = {};
vi E[maxn];
ld DP[maxn] = {};

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	cin >> N >> M;
	REP(i, N) cin >> C[i];

	REP(i, M) {
		int s, t;
		cin >> s >> t;
		E[t].pb(s);
		D[s]++;
	}

	queue<int> q;
	REP(i, N) if (D[i] == 0) q.push(i);

	while (!q.empty()) {
		int cur = q.front();
		q.pop();

		// find the best

		// Variant 1: you have to pick the current pizza:
		//DP[cur] = ((ld) 0.5) * DP[cur] + C[cur];
		// cerr << "DP[" << cur << "] = " << DP[cur] << endl;

		// Variant 2: you can choose to pick the current pizza:
		DP[cur] = max(DP[cur], ((ld) 0.5) * DP[cur] + C[cur]);

		for (int nxt : E[cur]) {
			assert(D[nxt] > 0);
			DP[nxt] = max(DP[nxt], DP[cur]);
			if (--D[nxt] == 0) {
				q.push(nxt);
			}
		}
	}

	ld answer = DP[0];

	(cout << fixed).precision(15);
	cout << answer << endl;
	return 0;
}