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:
# ABC152F

import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")



from collections import defaultdict
from queue import deque
import itertools
import functools
def find(start,goal):
    q = deque([start])
    seen = [None] * n
    seen[start] = -1
    while q:
        u = q.pop()
        d = seen[u]
        if u==goal:
            break
        for v in ns[u]:
            if seen[v] is None:
                seen[v] = u
                q.appendleft(v)
    u = goal
    es = 0
    while True:
        t = (min(u, seen[u]), max(u, seen[u]))
        es |= 1<<uv2e[u,seen[u]]
        if seen[u]==start:
            break
        u = seen[u]
    return es

n = int(input())
ns = defaultdict(set)
uv2e = {}
for i in range(n-1):
    u,v = map(int, input().split())
    ns[u-1].add(v-1)
    ns[v-1].add(u-1)
    uv2e[u-1,v-1] = i
    uv2e[v-1, u-1] = i
ans = 0
m = int(input())
paths = [None] * m
for i in range(m):
    u, v = map(int, input().split())
    paths[i] = find(u-1,v-1)

ans = 0

# def dfs(ind, s, ans):
#     if len(ind)==m:
#         if True in ind:
#             tmp = 0
#             for i in range(n-1):
#                 tmp += (s & 1<<i)>0
#             ans += 2**(n-1-tmp) * (-1)**((sum(ind)+1)%2)
#         return ans
#     ans = dfs(ind + [True], s | paths[len(ind)], ans)
#     ans = dfs(ind + [False], s, ans)
#     return ans
        
def bitlen(b):
    tmp = 0
    for i in range(n-1):
        tmp += (b&1<<i)>0
    return tmp
for ind in itertools.product(*[(False, True) for _ in range(m)]):
    num = sum(ind)+1
    if num==1:
        continue
    c = 2**(n - 1 - bitlen(functools.reduce(lambda x,y: x|y, (paths[i] for i in range(m) if ind[i]), 0)))
    ans += (-1)**num * c
# ans = dfs([], 0, 0)
print(2**(n-1) - ans)