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:
from itertools import combinations
import sys

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines


n = int(readline())
AB = list(list(map(int, readline().split())) for _ in range(n - 1))
m = int(readline())
UV = list(list(map(int, readline().split())) for _ in range(m))

graph = [[] for _ in range(n + 1)]
for i, (a, b) in enumerate(AB):
    graph[a].append((b, i))
    graph[b].append((a, i))


def dfs(s, g):
    kouho = 0
    par = [(0, 0)] * (n + 1)
    stack = []
    stack.append(s)
    while stack:
        v = stack.pop()
        if v == g:
            break
        for u, i in graph[v]:
            if u == par[v][0]:
                continue
            par[u] = (v, i)
            stack.append(u)
    # 経路復元
    p = g
    q, i = par[p]
    while q > 0:
        kouho |= 1 << i
        p = q
        q, i = par[q]
    return kouho


def popcnt(n):
    c = (n & 0x5555555555555555) + ((n >> 1) & 0x5555555555555555)
    c = (c & 0x3333333333333333) + ((c >> 2) & 0x3333333333333333)
    c = (c & 0x0F0F0F0F0F0F0F0F) + ((c >> 4) & 0x0F0F0F0F0F0F0F0F)
    c = (c & 0x00FF00FF00FF00FF) + ((c >> 8) & 0x00FF00FF00FF00FF)
    c = (c & 0x0000FFFF0000FFFF) + ((c >> 16) & 0x0000FFFF0000FFFF)
    c = (c & 0x00000000FFFFFFFF) + ((c >> 32) & 0x00000000FFFFFFFF)
    return c


sets = [0 for _ in range(m)]

for idx, (u, v) in enumerate(UV):
    sets[idx] = dfs(u, v)
ans = 0
for i in range(1, m + 1):
    kumiawase = list(combinations(range(m), i))
    for kumi in kumiawase:
        temp = 0
        for j in range(i):
            temp |= sets[kumi[j]]
        num = popcnt(temp)
        ans += (-1) ** (i + 1) * (pow(2, n - 1) - pow(2, n - 1 - num))

print(ans)