Problem: We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i.
Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?

* The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.

Constraints

* 2 \leq N \leq 50
* 1 \leq a_i,b_i \leq N
* The graph given in input is a tree.
* 1 \leq M \leq \min(20,\frac{N(N-1)}{2})
* 1 \leq u_i < v_i \leq N
* If i \not= j, either u_i \not=u_j or v_i\not=v_j
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M


Output

Print the number of ways to paint the edges that satisfy all of the M conditions.

Examples

Input

3
1 2
2 3
1
1 3


Output

3


Input

2
1 2
1
1 2


Output

1


Input

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5


Output

9


Input

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


Output

62

Solution (in PYTHON3):

Solution:
import sys
sys.setrecursionlimit(10**7) #再帰関数の上限,10**5以上の場合python
import math
from copy import copy, deepcopy
from copy import deepcopy as dcp
from operator import itemgetter
from bisect import bisect_left, bisect, bisect_right#2分探索
#bisect_left(l,x), bisect(l,x)#aはソート済みである必要あり。aの中からx未満の要素数を返す。rightだと以下
from collections import deque
#deque(l), pop(), append(x), popleft(), appendleft(x)
##listでqueの代用をするとO(N)の計算量がかかってしまうので注意
from collections import Counter#文字列を個数カウント辞書に、
#S=Counter(l),S.most_common(x),S.keys(),S.values(),S.items()
from itertools import accumulate,combinations,permutations#累積和
#list(accumulate(l))
from heapq import heapify,heappop,heappush
#heapify(q),heappush(q,a),heappop(q) #q=heapify(q)としないこと、返り値はNone
#import fractions#古いatcoderコンテストの場合GCDなどはここからimportする
from functools import lru_cache#pypyでもうごく
#@lru_cache(maxsize = None)#maxsizeは保存するデータ数の最大値、2**nが最も高効率
from decimal import Decimal

def input(): 
    x=sys.stdin.readline()
    return x[:-1] if x[-1]=="\n" else x
def printl(li): _=print(*li, sep="\n") if li else None
def argsort(s, return_sorted=False): 
    inds=sorted(range(len(s)), key=lambda k: s[k])
    if return_sorted: return inds, [s[i] for i in inds]
    return inds
def alp2num(c,cap=False): return ord(c)-97 if not cap else ord(c)-65
def num2alp(i,cap=False): return chr(i+97) if not cap else chr(i+65)
def matmat(A,B):
    K,N,M=len(B),len(A),len(B[0])
    return [[sum([(A[i][k]*B[k][j]) for k in range(K)]) for j in range(M)] for i in range(N)]
def matvec(M,v):
    N,size=len(v),len(M)
    return [sum([M[i][j]*v[j] for j in range(N)]) for i in range(size)]
def T(M):
    n,m=len(M),len(M[0])
    return [[M[j][i] for j in range(n)] for i in range(m)]
def main():
    mod = 1000000007
    #w.sort(key=itemgetter(1),reversed=True)  #二個目の要素で降順並び替え

    N = int(input())
    #N, K = map(int, input().split())
    #A = tuple(map(int, input().split())) #1行ベクトル
    #L = tuple(int(input()) for i in range(N)) #改行ベクトル
    #S = tuple(tuple(map(int, input().split())) for i in range(N)) #改行行列


    def tree_init(N):
        edgelist = [[] for _ in range(N)]
        for i, (a, b) in enumerate([map(int, input().split()) for i in range(N-1)]):#木じゃない場合N-1をNに変える
            edgelist[a-1].append((b-1,i))#ノードの番号が0から始まる場合-1を消す
            edgelist[b-1].append((a-1,i))
        return edgelist
    edge=tree_init(N)
    M = int(input())
    ms= tuple(tuple(map(int, input().split())) for i in range(M))
    ways=[]

    for u,v in ms:
        u-=1;v-=1
        q=deque([(u,[])])
        visited=[False]*N
        while q:
            x=q.pop()#ここをpopleftにすると幅優先探索BFSになる
            #print(x)
            e,way=x
            if visited[e]:continue
            #print(way)
            if e==v:
                ways.append(set(way))
            visited[e]=True
            for ne, j in edge[e]:
                nw=copy(way)
                nw.append(j)
                q.append((ne,nw))
                
    tot=pow(2,N-1)
    for i in range(1,M+1):
        for ws in combinations(ways,i):
            ts=set()
            for w in ws:
                ts|=w
            # print(len(ts))
            tot+=(-1)**i*pow(2,N-1-len(ts))
    print(tot)
    #print(2**49-1)
if __name__ == "__main__":
    main()