#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include <algorithm>
#include <atcoder/all>
#include <bitset>
#include <cassert>
#include <cmath>
#include <ctime>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using namespace atcoder;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define repk(i, k, n) for (int i = k; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define mod1 1000000007
#define mod2 998244353
#define mod3 100000007
#define vi vector<int>
#define vs vector<string>
#define vc vector<char>
#define vl vector<ll>
#define vb vector<bool>
#define vvi vector<vector<int>>
#define vvc vector<vector<char>>
#define vvl vector<vector<ll>>
#define vvb vector<vector<bool>>
#define vvvi vector<vector<vector<int>>>
#define vvvl vector<vector<vector<ll>>>
#define pii pair<int, int>
#define pil pair<int, ll>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define vpii vector<pair<int, int>>
#define vpll vector<pair<ll, ll>>
#define vvpii vector<vector<pair<int, int>>>
#define vvpll vector<vector<pair<ll, ll>>>

// using mint = modint998244353;

template <typename T>
void debug(T e) {
    cerr << e << endl;
}

template <typename T>
void debug(vector<T> &v) {
    rep(i, v.size()) { cerr << v[i] << " "; }
    cerr << endl;
}

template <typename T>
void debug(vector<vector<T>> &v) {
    rep(i, v.size()) {
        rep(j, v[i].size()) { cerr << v[i][j] << " "; }
        cerr << endl;
    }
}

template <typename T>
void debug(vector<pair<T, T>> &v) {
    rep(i, v.size()) { cerr << v[i].first << " " << v[i].second << endl; }
}

template <typename T>
void debug(set<T> &st) {
    for (auto itr = st.begin(); itr != st.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(multiset<T> &ms) {
    for (auto itr = ms.begin(); itr != ms.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(map<T, T> &mp) {
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        cerr << itr->first << " " << itr->second << endl;
    }
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << H << " ";
    debug_out(T...);
}

void bellman_ford(vector<ll> &from, vector<ll> &to, vector<ll> &cost,
                  vector<ll> &dis, ll v, ll N, ll M, bool &negative) {
    // ベルマンフォード法。
    // 頂点fromから頂点toまでの長さcostの辺が，M本張られている。頂点数はN。
    // このもとで，始点vからの最短距離disを求めることにする。
    // 負の閉路がある場合はnegativeがtrueになる。
    vector<bool> hasroute(N, false);
    hasroute[v] = true;
    vector<bool> nega(N, false);
    for (ll i = 0; i < N; i++) {
        bool update = false;
        for (ll j = 0; j < M; j++) {
            if (dis[to[j]] > dis[from[j]] + cost[j]) {
                dis[to[j]] = dis[from[j]] + cost[j];
                if (hasroute[from[j]]) hasroute[to[j]] = true;
                update = true;
                if (i == N - 1) {
                    nega[to[j]] = true;
                }
            }
        }
        if (!update) {
            break;
        }
    }

    for (ll i = 0; i < N; i++) {
        for (ll j = 0; j < M; j++) {
            if (hasroute[from[j]] && nega[from[j]]) nega[to[j]] = true;
        }
    }
    if (hasroute[N - 1] && nega[N - 1]) negative = true;
}

int main() {
    ll N, M;
    cin >> N >> M;
    vector<ll> A(N);
    for (ll i = 0; i < N; i++) {
        cin >> A[i];
    }
    vector<ll> a(M);
    vector<ll> b(M);
    vector<ll> c(M);
    for (ll i = 0; i < M; i++) {
        cin >> a[i] >> b[i] >> c[i];
        a[i]--;
        b[i]--;
    }
    vector<ll> from(N + M);
    vector<ll> to(N + M);
    vector<ll> cost(N + M);
    for (ll i = 0; i < N; i++) {
        from[i] = i;
        to[i] = N + i;
        cost[i] = -A[i];
    }
    ll INF = 2000000000000000003;
    for (ll i = 0; i < M; i++) {
        from[i + N] = N + a[i];
        to[i + N] = b[i];
        cost[i + N] = c[i];
    }
    bool negative = false;
    vector<ll> dis(2 * N, INF);
    dis[0] = 0;
    bellman_ford(from, to, cost, dis, 0, 2 * N, N + M, negative);
    if (negative) {
        cout << "inf" << endl;
    } else {
        cout << -dis[2 * N - 1] << endl;
    }
}