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
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
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
    
m = int(input())
uvs = [None]*m
for i in range(m):
    u,v = map(int, input().split())
    u -= 1
    v -= 1
    uvs[i] = (u,v)
    
from queue import deque    
def bfs(start):
    q = deque([start])
    seen = [None] * n
    seen[start] = 0
    while q:
        u = q.pop()
        d = seen[u]
        for v in ns[u]:
            if seen[v] is None:
                seen[v] = u
                q.appendleft(v)
    return seen

paths = [None]*m
for i,(u,v) in enumerate(uvs):
    seen = bfs(u)
    ss = set()
    z = v
    while z!=u:
        z1 = seen[z]
        ss.add(uv2e[z,z1])
        z = z1
    paths[i] = ss
    
ans = 0
def dfs(l, s):
    if len(l)==m:
        global ans
        val = pow(2, n-1-len(s)) * pow(-1, sum(l))
        ans += val
#         print(val, sum(l), ans)
        return
    l.append(0)
    dfs(l, s)
    l.pop()
    
    i = len(l)
    p = paths[i]
    l.append(1)
    dfs(l, s|p)
    l.pop()
    
dfs([], set())
# ans = pow(2,n-1) - ans
print(ans)