#!/usr/bin/pypy3
from collections import deque

Yes = "Yes"
Unknown = "Unknown"
No = "No"

def main():
    # 入力受取
    N, M = map(int, input().split())
    s, t, k = map(int, input().split())
    s -= 1
    t -= 1

    adj = [[] for _ in range(N)]
    for i in range(M):
        u, v = map(int, input().split())
        u -= 1
        v -= 1
        adj[u].append((v))
        adj[v].append((u))

    # BFS
    que = deque()
    dist = [0] * N
    visited = [False] * N

    que.append(s)
    dist[s] = 0
    visited[s] = True

    while len(que) != 0:
        v = que.popleft()

        for u in adj[v]:
            if not visited[u]:
                que.append(u)
                dist[u] = dist[v] + 1
                visited[u] = True

    # グラフ(V, F)に長さkの歩道が存在する
    if visited[t] and dist[t] <= k and dist[t] % 2 == k % 2 and len(adj[s]) != 0:
        print(Yes)
        return

    # グラフ(V, F)に長さkの歩道が存在しない

    # sとtが連結
    if visited[t]:
        # 最短経路長がkの偶奇と一致しない
        if dist[t] % 2 != k % 2:
            print("No")
            return

        # 最短経路長がkの偶奇と一致する
        # s == tのとき、N = 1かで判定
        if s == t:
            print(Unknown if N != 1 else No)
            return

        # s != tのとき、最短経路長を 1 または 2 にできる
        min_dist = (1 if k % 2 else 2)
        print(Unknown if min_dist <= k else No)
        return

    # sとtが非連結
    # N = 2 かつ k%2 == 0 のとき別処理
    if N == 2 and k % 2 == 0:
        print(No)
        return

    # それ以外のとき、最短経路長を 1 または 2 の好きな方にできる
    min_dist = (1 if k % 2 else 2)
    print(Unknown if min_dist <= k else No)
    return

if __name__ == "__main__":
    main()