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 defaultdict
import sys

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int1, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]

def main():
    def dfs(u, v, pu=-1):
        if u == v: return True
        for cu in to[u]:
            if cu == pu: continue
            if dfs(cu, v, u):
                path.append(uvtoi[u, cu])
                return True
        return False

    to = defaultdict(list)
    n = II()
    uvtoi = {}
    for i in range(n - 1):
        a, b = MI1()
        to[a].append(b)
        to[b].append(a)
        uvtoi[a, b] = i
        uvtoi[b, a] = i
    m = II()
    itoj = defaultdict(list)
    for j in range(m):
        u, v = MI1()
        path = []
        dfs(u, v)
        #print(path)
        for i in path: itoj[i].append(j)
    #print(itoj)
    en = len(itoj)
    dp = [[0] * (1 << m) for _ in range(en + 1)]
    dp[0][0] = 1
    for i, jj in enumerate(itoj.values()):
        mask = sum(1 << j for j in jj)
        for bit in range(1 << m):
            pre = dp[i][bit]
            if pre == 0: continue
            dp[i + 1][bit | mask] += pre
            dp[i + 1][bit] += pre
    #print(dp)
    ans=dp[-1][-1]*pow(2,n-1-en)
    print(ans)

main()
