#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = (ll)4e18;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
if (!(cin >> n >> m)) return 0;
vector<ll> a(n);
for (int i = 0; i < n; ++i) cin >> a[i];
vector<vector<int>> adj(n);
for (int i = 0; i < m; ++i) {
int u, v; cin >> u >> v; --u; --v;
adj[u].push_back(v);
adj[v].push_back(u);
}
auto solve_from = [&](int s) {
vector<ll> d(n, INF);
priority_queue<pair<ll,int>, vector<pair<ll,int>>, greater<pair<ll,int>>> pq;
d[s] = 0;
pq.emplace(0, s);
while (!pq.empty()) {
auto [dt, u] = pq.top(); pq.pop();
if (dt != d[u]) continue;
for (int v : adj[u]) {
// earliest time we can be at v = either move immediately (d[u] + 1),
// or wait (raise altitude) until we satisfy a[v]
ll nd = max(d[u] + 1, a[v]);
if (nd < d[v]) {
d[v] = nd;
pq.emplace(nd, v);
}
}
}
return d;
};
vector<ll> d1 = solve_from(0);
vector<ll> d2 = solve_from(n-1);
ll ans = INF;
// meet at node
for (int i = 0; i < n; ++i) {
if (d1[i] == INF || d2[i] == INF) continue;
ans = min(ans, 2 * max(d1[i], d2[i]));
}
// meet across an edge (one step offset)
for (int u = 0; u < n; ++u) {
for (int v : adj[u]) {
if (d1[u] == INF || d2[v] == INF) continue;
ans = min(ans, 2 * max(d1[u], d2[v]) + 1);
}
}
cout << ans << '\n';
return 0;
}
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