#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
//using mint = modint1000000007;
//const int mod = 1000000007;
//using mint = modint998244353;
//const int mod = 998244353;
//const int INF = 1e9;
//const long long LINF = 1e18;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i,l,r)for(int i=(l);i<(r);++i)
#define rrep(i, n) for (int i = (n) - 1; i >= 0; --i)
#define rrep2(i,l,r)for(int i=(r) - 1;i>=(l);--i)
#define all(x) (x).begin(),(x).end()
#define allR(x) (x).rbegin(),(x).rend()
#define P pair<int,int>
template<typename A, typename B> inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; }
template<typename A, typename B> inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; }
#include <vector>
const long long LINF = 1e18;

template<typename T>
class Bellman_ford {
public:
	Bellman_ford(int n, bool longestRoot = false) :n(n), longestRoot(longestRoot) {}
	void addEdge(int v, int w, T c) {
		g.push_back({ v, w, c *(longestRoot ? -1 : 1) });
	}
	std::vector<T> solve() {
		std::vector<T>d(n, LINF);
		d[0] = 0;
		// 通常更新パート
		// n-1回最短経路問題を解くと負閉路がない場合の更新が終わる
		for (int i = 0; i < n - 1; ++i) {
			for (int j = 0; j < (int)g.size(); ++j) {
				if (LINF == d[g[j].from]) continue;
				if (d[g[j].from] + g[j].cost < d[g[j].to]) d[g[j].to] = d[g[j].from] + g[j].cost;
			}
		}
		// 閉路の伝達パート
		std::vector<bool>update(n);
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < g.size(); ++j) {
				if (LINF == d[g[j].from]) continue;
				if (d[g[j].from] + g[j].cost < d[g[j].to]) update[g[j].to] = true;
				if (update[g[j].from]) update[g[j].to] = true;
			}
		}
		// 負の閉路が存在するなら更新
		for (int i = 0; i < n; ++i) {
			if (update[i])d[i] = -LINF;
			if (longestRoot) d[i] *= -1;
		}
		return d;
	}
private:
	struct Edge {
		int from, to;
		T cost;
		Edge(int _from, int _to, T _cost) :from(_from), to(_to), cost(_cost) {}
	};
	int n;
	bool longestRoot;
	std::vector<Edge>g;
};
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m; cin >> n >> m;
	vector<int>a(n);
	rep(i, n)cin >> a[i];
	Bellman_ford<long long> bf(n, true);
	rep(i, m) {
		int x, y, z; cin >> x >> y >> z;
		x--, y--;
		bf.addEdge(x, y, a[y] - z);
	}
	auto vec = bf.solve();
	if (LINF == vec.back())cout << "inf" << endl;
	else cout << a[0] + vec.back() << endl;
	return 0;
}