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
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())

N = I()
dist = [[100]*(N+1) for _ in range(N+1)]  # 2点間の最短距離
edges = {}

for i in range(N-1):
    a,b = MI()
    dist[a][b] = 1
    dist[b][a] = 1
    edges[(a,b)] = i
    edges[(b,a)] = i

for i in range(1,N+1):
    dist[i][i] = 0

for k in range(1,N+1):
    for i in range(1,N+1):
        for j in range(1,N+1):
            dist[i][j] = min(dist[i][j],dist[i][k]+dist[k][j])

M = I()
A = []  # uとvを結ぶ経路上にある辺の集合
for _ in range(M):
    u,v = MI()
    B = set()
    while True:
        if dist[u][v] == 1:
            B.add(edges[(u,v)])
            break
        else:
            for i in range(1,N+1):
                if dist[u][i] == dist[u][v]-1 and dist[i][v] == 1:
                    B.add(edges[(i,v)])
                    v = i
                    break
    A.append(B)


def popcount(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


# 包除原理

ans = 0
for i in range(2**M):
    C = set()
    for j in range(M):
        if (i >> j) & 1:
            C |= A[j]
    l = len(C)
    a = popcount(i)
    a %= 2
    ans += (-1)**a * 2**(N-1-l)

print(ans)