#include <bits/stdc++.h>
#define For(i, a, b) for(long long i = a; i < b; i++)
#define rep(i, n) For(i, 0, n)
#define rFor(i, a, b) for(long long i = a; i >= b; i--)
#define ALL(v) (v).begin(), (v).end()
#define rALL(v) (v).rbegin(), (v).rend()
using namespace std;

using lint = long long;
using ld = long double;

int INF = 2000000000;
lint LINF = 1000000000000000000;

template <class T>
struct Edge {
    int from, to;
    T cost;
    int idx;
    
    Edge() {}
    Edge(int to_) : to(to_) {}
    Edge(int to_, T cost_) : to(to_), cost(cost_) {}
    Edge(int from_, int to_, int idx_) : from(from_), to(to_), idx(idx_) {}
    Edge(int from_, int to_, T cost_, int idx_) : from(from_), to(to_), cost(cost_), idx(idx_) {}
};

template <class T> using Graph = vector<vector<Edge<T>>>;
using graph = Graph<long long>;
using edge = Edge<long long>;

#define add emplace_back

int main() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<int> x(k);
    rep(i, k) {
        cin >> x[i];
        x[i]--;
    }
    Graph<int> g(n);
    rep(_, m) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        g[u].add(v);
        g[v].add(u);
    }
    // cnt[i][j] := 頂点 i へ j 回の移動で到達できる人の数
    vector<vector<int>> cnt(n, vector<int>(20));
    rep(i, k) {
        // dp[s][v] := 通った頂点集合が s で、今 v にいることは可能か？
        vector<vector<bool>> dp(1 << n, vector<bool>(n, false));
        dp[1 << x[i]][x[i]] = true;
        rep(s, 1 << n) {
            rep(v, n) {
                for (auto e : g[v]) {
                    int nv = e.to;
                    if (s & (1 << nv)) {
                        continue;
                    }
                    int ns = s | (1 << nv);
                    if (dp[s][v]) {
                        dp[ns][nv] = true;
                    }
                }
            }
        }
        rep(s, 1 << n) {
            rep(v, n) {
                if (dp[s][v]) {
                    cnt[v][__builtin_popcount(s) - 1] |= (1 << i);
                }
            }
        }
    }
    rep(i, n) {
        rep(j, 20) {
            if (cnt[i][j] == (1 << k) - 1) {
                cout << "Yes" << endl;
                return 0;
            }
        }
    }
    cout << "No" << endl;
}