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:
import sys

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  # ignore trailing spaces

n = ni()
g = [[] for _ in range(n)]
for i in range(n-1):
    a, b = na()
    g[a-1].append(b-1)
    g[b-1].append(a-1)

def dfs(cur, pre, g, par, od):
    par[cur] = pre
    od.append(cur)
    for e in g[cur]:
        if e == pre: continue
        dfs(e, cur, g, par, od)

par = [-1] * n
od = []
dfs(0, -1, g, par, od)

def lca(a, b, par):
    ps = set()
    q = a
    while q != -1:
        ps.add(q)
        q = par[q]
    while b != -1:
        if b in ps:
            return b
        b = par[b]
    raise

m = ni()
cons = [0] * n
for i in range(m):
    u, v = na()
    u -= 1
    v -= 1
    lc = lca(u, v, par)
    while u != lc:
        cons[u] |= 1<<i
        u = par[u]
    while v != lc:
        cons[v] |= 1<<i
        v = par[v]

def st(b, m):
    # b = copy(a)
    for i in range(m):
        for j in range(1<<m):
            if (j>>i&1) == 0:
                b[j|1<<i] += b[j]
    return b

def ist(b, m):
    # b = copy(a)
    for i in range(m):
        for j in range(1<<m):
            if (j>>i&1) == 0:
                b[j|1<<i] -= b[j]
    return b

dp = [1] * (1<<m)
for cur in range(1,n):
    for i in range(1<<m):
        if (i&cons[cur]) == cons[cur]:
            dp[i] *= 2

# dp = [None for _ in range(n)]
# for i in range(n-1,-1,-1):
#     cur = od[i]
#     my = [0] * (1<<m)
#     my[0] = 1
#     if cur != 0:
#         my[cons[cur]] += 1
#     my = st(my, m)
#
#     for e in g[cur]:
#         if e == par[cur]: continue
#         for j in range(1<<m):
#             my[j] = my[j] * dp[e][j]
#     dp[i] = my

    # dp[i] = ist(my, m)

dp = ist(dp, m)

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