Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

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

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

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

Sample Output 4

62

The tree in this input is shown below:

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from collections import Counter,defaultdict,deque
import sys
import bisect
import math
import itertools
import string
import queue
import copy
# import numpy as np
# import scipy
from itertools import permutations, combinations
from heapq import heappop, heappush
# input = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7
 
def inp(): # n=1
    return int(input())
def inpm(): # x=1,y=2
    return map(int,input().split())
def inpl(): # a=[1,2,3,4,5,...,n]
    return list(map(int, input().split()))
def inpls(): # a=['1','2','3',...,'n']
    return list(input().split())
def inplm(n): # x=[] 複数行
    return list(int(input()) for _ in range(n))
def inpll(n): # [[1,1,1,1],[2,2,2,2],[3,3,3,3]]
    return [list(map(int, input().split())) for _ in range(n)]

def bfs(s,n,g,t): # 木構造の距離の算出 start, number, graph
    went = [False]*n
    dis = [0]*n
    que = deque([])
    que.append(s)
    went[s] = True
    while que:
        go = que.pop()
        for i in range(len(g[go])):
            if went[g[go][i]]:
                continue
            dis[g[go][i]] = dis[go] + 1
            went[g[go][i]] = True
            que.append(g[go][i])
    root = []
    root.append(t)
    x = dis[t]
    que1 = deque([])
    que1.append(t)
    while que1:
        go = que1.pop()
        for i in range(len(g[go])):
            if dis[g[go][i]] == x-1:
                que1.append(g[go][i])
                root.append(g[go][i])
                x -=1
                break
        if x == 0:
            break
    return root


def main():
    n = inp()
    g=[[] for _ in range(n)]
    path = defaultdict(int)
    for i in range(n-1):
        a,b = inpm()
        a -= 1
        b -= 1
        key = [str(a), ' ', str(b)]
        key = ''.join(key)        
        path[key] = i+1
        key = [str(b), ' ', str(a)]
        key = ''.join(key)     
        path[key] = i+1
        g[b].append(a)
        g[a].append(b)
    m = inp()
    ans = 0
    pre_root = [[] for _ in range(m)]
    for j in range(m):
        u,v = inpm()
        u -= 1
        v -= 1
        root1 = bfs(u,n,g,v)
        for i in range(len(root1)-1):
            a,b = root1[i],root1[i+1]
            key = [str(a), ' ', str(b)]
            key = ''.join(key)     
            x = path[key]
            pre_root[j].append(x)
    for k in range(pow(2,n-1)):
        flag = 1
        for i in range(m):
            flag1 = 0
            for j in range(len(pre_root[i])):
                go = pre_root[i][j]-1
                if (k>>go) & 1 == 1:
                    flag1 = 1
                    break
            if not flag1:
                flag = 0
                break           
        if flag:
            ans += 1
    print(ans)

if __name__ == "__main__":
    main()