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
import sys
input = sys.stdin.buffer.readline

N = int(input())
AB = [list(map(int, input().split())) for _ in range(N - 1)]
M = int(input())
UV = [list(map(int, input().split())) for _ in range(M)]

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
    
graph = [[] for _ in range(N + 1)]
for a, b in AB:
  graph[a].append(b)
  graph[b].append(a)
  
def make_tree(graph, root) -> (list, list):
  INF = 10 ** 15
  dist = [INF] * len(graph)
  parent = [-1] * len(graph)
  stack = [root]
  dist[root] = 0
  parent[root] = root
  while stack:
    s = stack.pop()
    for g in graph[s]:
      if dist[g] > dist[s] + 1:
        dist[g] = dist[s] + 1
        parent[g] = s
        stack.append(g)
  return dist, parent
 
def lca(u, v, dist, parent):
  distu, distv = dist[u], dist[v]
  if distu > distv:
    for _ in range(distu - distv):
      u = parent[u]
  if distv > distu:
    for _ in range(distv - distu):
      v = parent[v]
  while u != v:
    u, v = parent[u], parent[v]
  return u

dist, parent = make_tree(graph, 1)

tree_lca = []
for i in range(M):
    tmp = 0
    u, v = UV[i]
    l = lca(u, v, dist, parent)
    while u != l:
        tmp |= 1 << u
        u = parent[u]
    while v != l:
        tmp |= 1 << v
        v = parent[v]
    tree_lca.append(tmp)

answer = 2 ** (N - 1)
for i in range(1, 2 ** M):
    tmp = 0
    for j in range(M):
        if (i >> j) & 1:
            tmp |= tree_lca[j]
    cnt = popcnt(tmp)

    if popcnt(i) % 2 == 1:
        answer -= 2 ** (N - 1 - cnt) 
    else:
        answer += 2 ** (N - 1 - cnt) 
        
print(answer)