#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
using mint = atcoder::modint998244353;
// using mint = double;

#define rep(i, l, r) for (int i = (int)(l); i<(int)(r); i++)
#define ll long long
#define ld long double
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define siz(x) (int)(x).size()

template<class T> bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

const int inf = 1e9;
const ll INF = 4e18;

template<class T> using pq = priority_queue<T, vector<T>, less<T>>;
template<class T> using spq = priority_queue<T, vector<T>, greater<T>>;

vector<int> di = {0, 0, 1, -1};
vector<int> dj = {1, -1, 0, 0};

struct Edge {
    int to, cost;
};

//1次元累積和
//全て0-indexedで処理
template<class T>
struct CumulativeSum {
    int siz;
    vector<T> S;
    bool done;
    CumulativeSum() : CumulativeSum(0) {}
    CumulativeSum(int N) : CumulativeSum(vector<T>(N, 0)) {}
    //累積和の構築はしない
    CumulativeSum(vector<T> A) {
        done = false;
        siz = (int)A.size();
        S.resize(siz+1);
        S[0] = 0;
        for (int i = 0; i < siz; i++) {
            S[i+1] = A[i];
        }
    }
    //累積
    void build() {
        assert(!done);
        for (int i = 1; i <= siz; i++) {
            S[i] += S[i-1];
        }
        done = true;
    }
    //加算（累積前のみ）
    T add(int idx, T a) {
        assert(!done);
        return S[idx+1] += a;
    }
    //代入（累積前のみ）
    T set(int idx, T a) {
        assert(!done);
        return S[idx+1] = a;
    }
    //取得（累積前、累積後両方とも可能だが、どちらを想定するかをexpected_doneで渡す）
    [[nodiscard]]
    T get(int idx, bool expected_done) {
        assert(expected_done == done);
        return S[idx+1];
    }
    //区間和取得（累積後のみ）
    //半開区間で与える
    [[nodiscard]]
    T sum(int l, int r) {
        assert(done);
        return S[r]-S[l];
    }
};

void solve() {
    int N, M, S, T; ll K; cin >> N >> M >> S >> T;
    vector<int> deg(N, 0);
    rep(i, 0, M) {
        int u, v; cin >> u >> v;
        u--; v--;
        deg[u]++; deg[v]++;
    }
    if (K%2) {
        cout << "Yes" << endl;
    }
    else {
        if (deg[S] or deg[T]) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int T = 1;
    // cin >> T;
    while(T--) {
        solve();
    }
}