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
#
# 　　  ⋀_⋀　 
#　　  (･ω･)  
# .／ Ｕ ∽ Ｕ＼
#  │＊　合　＊│
#  │＊　格　＊│ 
#  │＊　祈　＊│ 
#  │＊　願　＊│ 
#  │＊　　　＊│ 
#      ￣
#
import sys
sys.setrecursionlimit(10**6)
input=sys.stdin.readline
from math import floor,sqrt,factorial,hypot,log #log2ないｙｐ
from heapq import heappop, heappush, heappushpop
from collections import Counter,defaultdict,deque
from itertools import accumulate,permutations,combinations,product,combinations_with_replacement
from bisect import bisect_left,bisect_right
from copy import deepcopy
from fractions import gcd
from random import randint
def ceil(a,b): return (a+b-1)//b
inf=float('inf')
mod = 10**9+7
def pprint(*A): 
    for a in A:     print(*a,sep='\n')
def INT_(n): return int(n)-1
def MI(): return map(int,input().split())
def MF(): return map(float, input().split())
def MI_(): return map(INT_,input().split())
def LI(): return list(MI())
def LI_(): return [int(x) - 1 for x in input().split()]
def LF(): return list(MF())
def LIN(n:int): return [I() for _ in range(n)]
def LLIN(n: int): return [LI() for _ in range(n)]
def LLIN_(n: int): return [LI_() for _ in range(n)]
def LLI(): return [list(map(int, l.split() )) for l in input()]
def I(): return int(input())
def F(): return float(input())
def ST(): return input().replace('\n', '')
def main():
    N=I()
    edge = [[] for _ in range(N)]
    for i in range(N-1):
        a,b = MI_()
        edge[a].append(b)
        edge[b].append(a)
    
    root = 0
    q = []
    q.append(root)
    pre = [-1]*N
    visited = [False]*N
    while q:
        v = q.pop() 
        visited[v] = True
        for to in edge[v]:
            if visited[to]:
                continue
            pre[to] = v
            q.append(to)
    
    path = [] #パスに含まれる辺(根に向かう方で保存)
    constraint = []
    M=I()
    for i in range(M):
        a,b = MI_()
        constraint.append((a,b))

        
        root_path_a = 1<<a
        root_path_b = 1<<b
        
        while a!=root:
            root_path_a |= 1<<pre[a]
            a = pre[a]
        while b!=root:
            root_path_b |= 1<<pre[b]
            b = pre[b]
        path.append(root_path_a ^ root_path_b) #LCAまでのみ
        if not path[-1]>>a&1:
            path[-1] |= 1<<a
        if not path[-1]>>b&1:
            path[-1] |= 1<<b
    

    ans = 0
    for mask in range(1<<M):
        builtin_count = 0
        U = 0 #pathに含まれる辺
        for i in range(M):
            if mask>>i&1:
                builtin_count += 1
                U |= path[i]
        edge_cnt = max(0,bin(U).count("1")-1)
        if builtin_count&1:
            ans -= 1<<(N-1-edge_cnt)
        else:
            ans += 1<<(N-1-edge_cnt)
    print(ans)



if __name__ == '__main__':
    main()