Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

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

Sample Output 4

62

The tree in this input is shown below:

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import sys
stdin = sys.stdin
sys.setrecursionlimit(10**6)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()

# cook your dish here
def dijkstra_back(s,t,n,w,cost):
    #sからtへの最短経路の経路復元
    prev = [s] * n #最短経路の直前の頂点
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True
        
        for i in range(n):
            if d[i] > d[v] + cost[v][i]: 
                d[i] = d[v] + cost[v][i]
                prev[i] = v
        
    path = [t]
    while prev[t] != s:
        path.append(prev[t])
        prev[t] = prev[prev[t]]
    path.append(s)
    path = path[::-1]
    return path

################################

n = ni()
cost = [[float("inf") for i in range(n)] for i in range(n)] 
#cost[u][v] : 辺uvのコスト(存在しないときはinf この場合は10**10)
for i in range(n-1):
    a,b = na()
    cost[a-1][b-1] = 1
    cost[b-1][a-1] = 1

dp = set()
m = ni()
for i in range(m):
    u,v = na()
    d = dijkstra_back(u-1,v-1,n,n-1,cost)
    for j in range(len(d)-1):
        a,b = min(d[j],d[j+1]),max(d[j],d[j+1])
        dp.add((a,b))

a = len(dp)
print(2**n-2**(n-1-a))