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:
from collections import deque

def bfs():
    q = deque([0])
    bit = [-1]*N
    bit[0] = 0
    serial = 0
    
    while q:
        v = q.popleft()
        
        for nv in adj_list[v]:
            if bit[nv]!=-1:
                continue
            bit[nv] = bit[v]|(1<<serial)
            serial += 1
            q.append(nv)
    
    return bit

N = int(input())
adj_list = [[] for _ in range(N)]

for _ in range(N-1):
    ai, bi = map(int, input().split())
    adj_list[ai-1].append(bi-1)
    adj_list[bi-1].append(ai-1)

bit = bfs()
M = int(input())
uv = []
path_bit = []

for _ in range(M):
    ui, vi = map(int, input().split())
    uv.append((ui-1, vi-1))
    path_bit.append(bit[ui-1]^bit[vi-1])

houzyo = 0

for i in range(1, 2**M):
    covered = 0
    c = 0
    
    for j in range(M):
        if (i>>j)&1:
            covered |= path_bit[j]
            c += 1
    
    no_covered = 0
    
    for j in range(N-1):
        if not (covered>>j)&1:
            no_covered += 1
    
    if c%2==1:
        houzyo += 2**no_covered
    else:
        houzyo -= 2**no_covered

print(2**(N-1)-houzyo)