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
import sys
sys.setrecursionlimit(10 ** 7)

def dfs(G, p, pre, t, edge, x):
    for nxt in G[p]:
        if nxt == pre:
            continue
        a, b = p, nxt
        if b < a:
            a, b = b, a
        if nxt == t:
            edge[(a, b)] += x
            return True
        if dfs(G, nxt, p, t, edge, x):
            edge[(a, b)] += x
            return True
    return False

def main():
    n = int(input())
    G = [[] for _ in range(n)]
    edge = {}
    for _ in range(n-1):
        a, b = map(int, input().split())
        a, b = a-1, b-1
        G[a].append(b)
        G[b].append(a)
        if b < a:
            a, b = b, a
        edge[(a, b)] = 0
    m = int(input())
    for i in range(m):
        u, v = map(int, input().split())
        u, v = u-1, v-1
        x = 1<<i
        dfs(G, u, -1, v, edge, x)
    ans = 2**(n-1)
    for i in range(1, 2**m):
        bit_cnt = 0
        for j in range(m):
            if (i & (1<<j)) != 0:
                bit_cnt += 1
        cnt = 0
        for v in edge.values():
            if (v & i) != 0:
                cnt += 1
        x = 2**(n-1 - cnt)
        if bit_cnt % 2 == 0:
            ans += x
        else:
            ans -= x
    print(ans)

if __name__ == "__main__":
    main()