Problem: 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

Solution (in PYTHON3):

Solution:
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)]

from collections import deque

links = [[] for _ in range(n+1)]
for ind,val in enumerate(ab):
    a,b = val
    links[a].append((ind,b))
    links[b].append((ind,a))

def bfs(start,end):
    parent = [[-1,-1] for _ in range(n+1)]
    d = deque()
    d.append(start)
    done = set()
    while(d):
        tmp = d.popleft()
        done.add(tmp)
        for i in links[tmp]:
            link_ind, j = i
            if(not j in done):
                parent[j] = [tmp,link_ind]
                d.append(j)
    route = []
    while(True):
        route.append(parent[end][1])
        end = parent[end][0]
        if(end == start):
            break
    return route[::-1]

links_key = [0] * (n-1)
for i,se in enumerate(uv):
    start,end = se
    route = bfs(start,end)
    for j in route:
        links_key[j] =(links_key[j] | 1<<i)

dp = [[0] * (2**m) for _ in range(n)]
dp[0][0] = 1

for i in range(n-1):
     for j in range(2**m):
         dp[i+1][j] += dp[i][j]
         dp[i+1][(j|links_key[i])] += dp[i][j]

print(dp[-1][-1])