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,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_str(): return list(sys.stdin.readline().split())

N = inp()
lines = [[] for _ in range(N)]
for i in range(N-1):
    a,b = inpl()
    a-=1; b-=1
    lines[a].append((b,i))
    lines[b].append((a,i))


def dfs(s,b,v):
    if s == v:
        return set()
    else:
        for t,i in lines[s]:
            if t == b:
                continue
            ret = dfs(t,s,v)
            if ret is not None:
                ret.add(i)
                return ret

        return None


M = inp()
mm = [] # i 番目の条件は mm[i] のラインのどこかに +1 をする
for m in range(M):
    u,v = inpl()
    u-=1; v-=1
    mm.append(dfs(u,-1,v))



dp = [0] * (1<<M)
dp[0] = 1

for i in range(N-1):
    # line i を白くした場合
    # dp[:] = dp[:]
    # line i を黒くした場合
    tmp = 0
    for b,m in enumerate(mm):
        if i in m:
            tmp += (1<<b)

    for b in reversed(range(1<<M)):
        dp[b | tmp] += dp[b]



print(dp[-1])
