#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 10000;
const int INF = 2147483647;
int n, m, b;     // 城市数量，公路数量，血量
int f[MAXN + 5]; // 每个城市过路费
// 存图 <终点,血量花费>
vector<pair<int, int>> e[MAXN + 5];

// 检查花费最大值小于等于 mid 时能否达成
int dis[MAXN + 5]; // 最少花费血量
bool vis[MAXN + 5];
// <花费血量,城市编号>
priority_queue<pair<int, int>> pq;
bool check(int mid)
{
    if (f[1] > mid)
        return false;
    for (int i = 1; i <= n; i++)
    {
        dis[i] = INF;
        vis[i] = false;
    }
    dis[1] = 0;
    pq.push(make_pair(-0, 1));
    while (!pq.empty())
    {
        int u = pq.top().second;
        pq.pop();
        if (vis[u])
            continue;
        vis[u] = true;
        for (int i = 0; i < e[u].size(); i++)
        {
            int v = e[u][i].first;
            int w = e[u][i].second;
            if (f[v] <= mid && dis[u] + w < dis[v])
            {
                dis[v] = dis[u] + w;
                pq.push(make_pair(-dis[v], v));
            }
        }
    }
    return dis[n] <= b;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m >> b;
    for (int i = 1; i <= n; i++)
        cin >> f[i];
    for (int i = 1; i <= m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        e[u].push_back(make_pair(v, w));
        e[v].push_back(make_pair(u, w));
    }

    // 二分限制花费的最大值
    int l = 0;
    int r = 1'000'000'000;
    int ans = -1;
    while (l <= r)
    {
        int mid = (l + r) / 2;
        if (check(mid))
        {
            ans = mid;
            r = mid - 1;
        }
        else
        {
            l = mid + 1;
        }
    }

    if (ans == -1)
        cout << "AFK";
    else
        cout << ans;
    return 0;
}