Problem: You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.

A tree is a connected undirected graph with n-1 edges.

You are given m queries. The i-th query consists of the set of k_i distinct vertices v_i[1], v_i[2], ..., v_i[k_i]. Your task is to say if there is a path from the root to some vertex u such that each of the given k vertices is either belongs to this path or has the distance 1 to some vertex of this path.

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 2 ⋅ 10^5) — the number of vertices in the tree and the number of queries.

Each of the next n-1 lines describes an edge of the tree. Edge i is denoted by two integers u_i and v_i, the labels of vertices it connects (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i).

It is guaranteed that the given edges form a tree.

The next m lines describe queries. The i-th line describes the i-th query and starts with the integer k_i (1 ≤ k_i ≤ n) — the number of vertices in the current query. Then k_i integers follow: v_i[1], v_i[2], ..., v_i[k_i] (1 ≤ v_i[j] ≤ n), where v_i[j] is the j-th vertex of the i-th query.

It is guaranteed that all vertices in a single query are distinct.

It is guaranteed that the sum of k_i does not exceed 2 ⋅ 10^5 (∑_{i=1}^{m} k_i ≤ 2 ⋅ 10^5).

Output

For each query, print the answer — "YES", if there is a path from the root to some vertex u such that each of the given k vertices is either belongs to this path or has the distance 1 to some vertex of this path and "NO" otherwise.

Example

Input


10 6
1 2
1 3
1 4
2 5
2 6
3 7
7 8
7 9
9 10
4 3 8 9 10
3 2 4 6
3 2 1 5
3 4 8 2
2 6 10
3 5 4 7


Output


YES
YES
YES
YES
NO
NO

Note

The picture corresponding to the example:

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Consider the queries.

The first query is [3, 8, 9, 10]. The answer is "YES" as you can choose the path from the root 1 to the vertex u=10. Then vertices [3, 9, 10] belong to the path from 1 to 10 and the vertex 8 has distance 1 to the vertex 7 which also belongs to this path.

The second query is [2, 4, 6]. The answer is "YES" as you can choose the path to the vertex u=2. Then the vertex 4 has distance 1 to the vertex 1 which belongs to this path and the vertex 6 has distance 1 to the vertex 2 which belongs to this path.

The third query is [2, 1, 5]. The answer is "YES" as you can choose the path to the vertex u=5 and all vertices of the query belong to this path.

The fourth query is [4, 8, 2]. The answer is "YES" as you can choose the path to the vertex u=9 so vertices 2 and 4 both have distance 1 to the vertex 1 which belongs to this path and the vertex 8 has distance 1 to the vertex 7 which belongs to this path.

The fifth and the sixth queries both have answer "NO" because you cannot choose suitable vertex u.

Solution (in PYTHON3):

Solution:
from sys import stdin, stdout
from collections import deque
from math import log
input = stdin.readline
print = stdout.write

n, m = map(int, input().split())
e = [[] for _ in range(n+1)]
for _ in range(n-1):
    a, b = map(int, input().split())
    e[a].append(b); e[b].append(a)

par, dep = [0]*(n+1), [0]*(n+1)
q = deque([(1, 1, 1)])
while q:
    p, d, cur = q.popleft()
    par[cur], dep[cur] = p, d
    for u in e[cur]:
        if u == p: continue
        q.append((cur, d+1, u))

nlg = int(log(max(dep), 2)) + 1
dp = [[0]*(n+1) for _ in range(nlg+1)]
for i in range(1, n+1): dp[0][i] = par[i]
for i in range(1, nlg+1):
    for j in range(1, n+1):
        dp[i][j] = dp[i-1][dp[i-1][j]]

def blift(u, jump):
    lg = 0
    while jump:
        if jump % 2: u = dp[lg][u]
        jump //= 2; lg += 1
    return u

for _ in range(m):
    ls = list(map(int, input().split()))
    sz, ls = ls[0], ls[1:]
    path = []
    for u in ls:
        p = par[u]
        path.append((dep[p], p))
    path = list(set(path))
    path.sort()

    ok = 1
    for i in range(1, len(path)):
        if path[i-1][0] == path[i][0]: ok = 0
    if not ok: print("NO\n"); continue

    for i in range(1, len(path)):
        d = path[i][0] - path[i-1][0]
        if path[i-1][1] != blift(path[i][1], d): ok = 0
    if not ok: print("NO\n"); continue

    print("YES\n")