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

def popcount(x):
    x = x - ((x >> 1) & 0x5555555555555555)
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
    x = x + (x >> 8)
    x = x + (x >> 16)
    x = x + (x >> 32)
    return x & 0x0000007f


def main():
    import sys
    import collections
    input = sys.stdin.readline
    N = int(input())
    G = [[-1]*N for i in range(N)]

    ans = 2**(N-1)
    for i in range(N-1):
        a, b = map(int, input().split())
        a -= 1
        b -= 1
        G[a][b] = i
        G[b][a] = i
    ans = 0
    g = [[0]*N for i in range(N)]
    for i in range(N):
        q = collections.deque()
        q.append((i, 0))
        r = [True]*N
        r[i] = False
        while(q):
            x, s = q.popleft()
            for y in range(N):
                if G[x][y] >= 0 and r[y]:
                    g[i][y] = s | 1 << G[x][y]
                    q.append((y, g[i][y]))
                    r[y] = False
    M = int(input())
    s = [0 for i in range(M)]
    for i in range(M):
        u, v = map(int, input().split())
        u -= 1
        v -= 1

        s[i] = g[u][v]

    # print(s)
    for i in range(2**M):
        tmp = 2**(N-1) - 1
        c = 0
        for j in range(M):
            if (i >> j) % 2 == 1:
                tmp &= ~s[j]
                c += 1
        ans += ((-1)**(c))*(1 << popcount(tmp))

    print(ans)


main()
