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:
# F - Tree and Constraints

# Read input

n = int(input())
adj = {a: [] for a in range(1, n + 1)}
for i in range(n - 1):
    a, b = map(int, input().split())
    adj[a].append((b, i))
    adj[b].append((a, i))
m = int(input())
ranges = [tuple(map(int, input().split())) for i in range(m)]

# Subroutines

def findpath(u, v, parent=None):
    """uからvへの経路 (辺の集合)"""
    if u == v:
        return set()
    for x, i in adj[u]:
        if x == parent:
            continue
        p = findpath(x, v, u)
        if p is not None:
            p.add(i)  # id of edge
            return p
    return None

def powerset(seq, bag):
    """[(seqの部分集合s, bag[e] for e in s の和集合)]"""
    if len(seq) <= 1:
        yield ((seq[0],), bag[seq[0]])
        yield ((), set())
    else:
        for s, b in powerset(seq[1:], bag):
            yield ((seq[0],) + s, b | bag[seq[0]])
            yield (s, b)

# Find paths

paths = [findpath(u, v) for u, v in ranges]

# いずれかの区間をすべて白にする塗り方の個数
white = 0
for s, b in powerset(list(range(m)), paths):
    l = len(s)
    if l <= 0:
        continue
    # b中の辺をすべて白にする塗り方の個数
    w = 2 ** (n - 1 - len(b))
    # 包除原理
    sign = -1 if l % 2 == 0 else 1    
    white += sign * w

# どの区間も黒を含む塗り方の個数
coloring = 2 ** (n - 1) - white
print(coloring)