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 main():
    input = sys.stdin.readline
    N = int(input())
    E = [[] for _ in range(N)]
    for i in range(N-1):
        a, b = map(int, input().split())
        a, b = a-1, b-1
        E[a].append((b,i))
        E[b].append((a,i))
    
    def dfs(s, g):
        par = [(-1,-1)] * N
        par[s] = (s,-1)
        stack = [s]
        while stack:
            v = stack.pop()
            for to, i in E[v]:
                if par[to][0] >= 0: continue
                par[to] = (v, i)
                if to == g: break
                stack.append(to)
        r = 0
        v = g
        while v != s:
            v, i = par[v]
            r |= 1 << i
        return r
    
    M = int(input())
    path = [0] * M
    for i in range(M):
        u, v = map(int, input().split())
        u, v = u-1, v-1
        path[i] = dfs(u, v)

    def calc(s): return 1<<(N-1-popcount(s))

    i_ans = 0
    for p in range(1, 1<<M):
        is_odd = 0
        s = 0
        for i in range(M):
            if p&1:
                s |= path[i]
                is_odd ^= 1
            p >>= 1
        if is_odd: i_ans += calc(s)
        else: i_ans -= calc(s)

    print((1<<(N-1)) - i_ans)

def popcount(x):
    x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
    x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
    x = (x & 0x0f0f0f0f0f0f0f0f) + (x >> 4 & 0x0f0f0f0f0f0f0f0f);
    x = (x & 0x00ff00ff00ff00ff) + (x >> 8 & 0x00ff00ff00ff00ff);
    x = (x & 0x0000ffff0000ffff) + (x >>16 & 0x0000ffff0000ffff);
    return (x & 0x00000000ffffffff) + (x >>32 & 0x00000000ffffffff);

if __name__ == '__main__':
    main()