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 *

def popcnt(n):
    c = (n & 0x5555555555555555) + ((n>>1) & 0x5555555555555555)
    c = (c & 0x3333333333333333) + ((c>>2) & 0x3333333333333333)
    c = (c & 0x0f0f0f0f0f0f0f0f) + ((c>>4) & 0x0f0f0f0f0f0f0f0f)
    c = (c & 0x00ff00ff00ff00ff) + ((c>>8) & 0x00ff00ff00ff00ff)
    c = (c & 0x0000ffff0000ffff) + ((c>>16) & 0x0000ffff0000ffff)
    c = (c & 0x00000000ffffffff) + ((c>>32) & 0x00000000ffffffff)
    return c

N = int(input())
edges = [[] for i in range(N)]
edge_id = defaultdict(lambda : -1)
for i in range(N-1):
    a, b = map(lambda x: int(x)-1, input().split())
    edges[a].append(b)
    edges[b].append(a)
    edge_id[(a, b)] = edge_id[(b, a)] = i
path = defaultdict(lambda : -1)

def set_path(r, v, path_):
    visited[v] = 1
    path[(r, v)] = path_
    for u in edges[v]:
        if visited[u] == 0:
            set_path(r, u, path_ | (1 << edge_id[(v, u)]))

for i in range(N):
    visited = [0] * N
    set_path(i, i, 0)
M = int(input())
uv = [list(map(lambda x: int(x)-1, input().split())) for i in range(M)]
ans = pow(2, N-1)
for i in range(1, 1 << M):
    all_path = 0
    for j, (u, v) in enumerate(uv):
        if i & (1 << j):
            all_path |= path[(u, v)]
    if popcnt(i) & 1:
        ans -= pow(2, N-1-popcnt(all_path))
    else:
        ans += pow(2, N-1-popcnt(all_path))
print(ans)
