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
#from collections import deque,defaultdict
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl   = lambda: list(map(int, input().split()))
inm   = lambda:      map(int, input().split())
ins = lambda : input().strip()
DBG = True  and False
BIG = 10**18
R = 10**9 + 7
#R = 998244353

def ddprint(x):
  if DBG:
    print(x)

def popcnt(x):
    c = 0
    while x>0:
        c += x%2
        x >>= 1
    return c

def dfs(u,v,p):
    global d
    d[u][v] = p
    for x in dst[v]:
        if u!=x and d[u][x]==0:
            p |= 1<<dst[v][x]
            dfs(u,x,p)
            p ^= 1<<dst[v][x]

n = inn()
dst = [{} for i in range(n+1)]
for i in range(n-1):
    a,b = inm()
    dst[a][b] = dst[b][a] = i

m = inn()
u = []
v = []
for i in range(m):
    uu,vv = inm()
    u.append(uu)
    v.append(vv)

if False and n==2 and m==1:
    print(0)
    exit()

d = [[0]*(n+1) for i in range(n+1)]
for i in range(1,n+1):
    dfs(i,i,0)
dd = []
for i in range(m):
    dd.append(d[u[i]][v[i]])
ddprint(dst)
ddprint(d)
ddprint(u)
ddprint(v)
ddprint(dd)

y = 0
for i in range(1,2**m):
    nbits = x = 0
    for j in range(m):
        if (i>>j)%2>0:
            x |= dd[j]
            nbits += 1
    nsegs = popcnt(x)
    s = 2*(nbits%2)-1
    y += s*2**(n-1-nsegs)
print(2**(n-1)-y)
