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:
# 解説AC
from collections import deque

N = int(input())
edge = [[] for _ in range(N)]
for n in range(N - 1):
    a,b = map(int, input().split())
    edge[a-1].append([b-1, n])
    edge[b-1].append([a-1, n])

def bfs(s:int, g:int):
    """ 始点sから終点gまでのパスを求める """
    in_edge = [-1] * N
    par = [-1] * N
    flag = False # 終点到着フラグ
    
    que = deque([s])
    while que:
        v = que.popleft()
        for w, n in edge[v]:
            if w == par[v]: continue
            in_edge[w] = n
            par[w] = v
            if w == g:
                flag = True
                break
            que.append(w)
        if flag: break
    path = 0
    cur = g
    while in_edge[cur] >= 0:
        path |= 1 << in_edge[cur]
        cur = par[cur]
    return path


M = int(input())
path = [0] * M # 各制約で使用するパス
for i in range(M):
    u,v = map(int, input().split())
    path[i] = bfs(u - 1, v - 1)

ans = 0
# 包除原理
for bit in range(1 << M):
    val = 0
    for i in range(M):
        if (bit >> i) & 1:
            val |= path[i]
    if bin(bit).count("1") % 2 == 0:
        ans += 1 << (N - 1 - bin(val).count("1"))
    else:
        ans -= 1 << (N - 1 - bin(val).count("1"))

print(ans)