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
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = 1e10

#solve
def solve():
    def dfs(p, g, pre=-1):
        for e in edg[p]:
            if e == pre: continue
            if e == g:
                return [d[(p, e)]]
            tmp = dfs(e, g, p)
            if tmp:
                return tmp + [d[(p, e)]]
        return False

    n = II()
    edg = [[] for i in range(n)]
    t = 0
    ab = defaultdict(int)
    d = defaultdict(int)
    for _ in range(n - 1):
        a, b = LI_()
        edg[a].append(b)
        edg[b].append(a)
        d[(a, b)] = t
        d[(b, a)] = t
        t += 1
    m = II()
    path = [0] * m 
    uv = LIR_(m)
    for i, uvi in enumerate(uv):
        u, v = uvi
        pat = dfs(u,v)
        for p in pat:
            path[i] |= 1 << p
    ans = 1 << (n - 1)
    for mask in range(1, 1 << m):
        tmp = 0
        t = 0
        for i in range(m):
            if mask & (1 << i):
                t += 1
                tmp = tmp | path[i]
        x = 0
        while tmp:  
            x += tmp & 1
            tmp >>=1
        ans += (1 << (n - x - 1)) * (-1) ** t
    print(ans)


    return


#main
if __name__ == '__main__':
    solve()
