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.readline

N = int(input())
graph = [[] for _ in range(N)]
Edges = [[None]*N for _ in range(N)]
for i in range(N-1):
    a, b = map(int, input().split())
    graph[a-1].append(b-1)
    graph[b-1].append(a-1)
    Edges[a-1][b-1] = i
    Edges[b-1][a-1] = i

def dfs(p, L, checked, goal):
    if p == goal:
        return L
    checked[p] = True
    ok = False
    for np in graph[p]:
        if not checked[np]:
            L.append(np)
            retL = dfs(np, L, checked, goal)
            if retL:
                ok = True
                break
            L.pop()
    if ok:
        return retL
    else:
        return []


M = int(input())
Compression = []
for _ in range(M):
    a, b = map(int, input().split())
    a -= 1; b -= 1
    path = dfs(a, [], [False]*N, b)
    bit = 0
    pre = a
    for p in path:
        bit += 1<<Edges[pre][p]
        pre = p
    Compression.append(bit)

ans = 0
for bitmask in range(1,1<<M):
    whitebit = 0
    for i in range(M):
        if (1<<i)&bitmask:
            whitebit |= Compression[i]
    
    p = 1
    for n in range(M):
        if bitmask&(1<<n):
            p ^= 1
    E = 1
    for m in range(N-1):
        if whitebit&(1<<m) == 0:
            E *= 2
    
    ans += E*(-1)**p

print(2**(N-1)-ans)