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
N = int(input())
E = [[int(l) for l in input().split()] for _ in range(N-1)]
M = int(input())
L = [[int(l) for l in input().split()] for _ in range(M)]

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.root = [-1]*(n+1)
        self.rnk = [0]*(n+1)
        
    def find_root(self, x):
        if self.root[x] < 0:
            return x
        else:
            self.root[x] = self.find_root(self.root[x])
            return self.root[x]
        
    def unite(self, x, y):
        x = self.find_root(x)
        y = self.find_root(y)
        if x == y:
            return
        elif self.rnk[x] > self.rnk[y]:
            self.root[x] += self.root[y]
            self.root[y] = x
        else:
            self.root[y] += self.root[x]
            self.root[x] = y
            if self.rnk[x] == self.rnk[y]:
                self.rnk[y] += 1
                
    def isSameGroup(self, x, y):
        return self.find_root(x) == self.find_root(y)

C = [0]*M
for i in range(N-1):
    uf = UnionFind(N)
    for e in E:
        if e == E[i]:
            continue
        uf.unite(e[0], e[1])
    for l in range(M):
        if uf.isSameGroup(L[l][0], L[l][1]):
            continue
        C[l] |= 1 << i

import itertools

ans = 0
L = [int(i) for i in range(M)]
for m in range(M+1):
    for P in itertools.combinations(L, m):
        white = 0
        for j in P:
            white |= C[j]
        if m%2:
            ans -= 1 << N-1-bin(white).count("1")
        else:
            ans += 1 << N-1-bin(white).count("1")
print(ans)