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
input = sys.stdin.readline
sys.setrecursionlimit(10**9)

def solve():
    N = int(input())
    adjL = [[] for _ in range(N)]
    edges = []
    for _ in range(N-1):
        a, b = map(int, input().split())
        a, b = a-1, b-1
        adjL[a].append(b)
        adjL[b].append(a)
        edges.append((a, b))
    M = int(input())
    uvs = [tuple(map(lambda x: int(x)-1, input().split())) for _ in range(M)]

    e2Roots = [set() for _ in range(N)]
    def dfs(vNow, vPar):
        for v2 in adjL[vNow]:
            if v2 == vPar: continue
            e2Roots[v2] |= e2Roots[vNow]
            e2Roots[v2].add((vNow, v2))
            dfs(v2, vNow)

    dfs(0, -1)

    ePaths = [set() for _ in range(M)]
    for i, (u, v) in enumerate(uvs):
        ePaths[i] = e2Roots[u] ^ e2Roots[v]

    numEs = [0] * (1<<M)
    for a, b in edges:
        ptn = 0
        for iC, ePath in enumerate(ePaths):
            if (a, b) in ePath or (b, a) in ePath:
                ptn |= 1<<iC
        for S in range(1<<M):
            if S & ptn:
                numEs[S] += 1

    def getPows(base, n):
        pows = [1] * (n+1)
        for x in range(1, n+1):
            pows[x] = (pows[x-1] * base)
        return pows
    pows = getPows(2, N)

    def getNum1s(N):
        num1s = [0]
        for _ in range(N):
            num12s = [num1+1 for num1 in num1s]
            num1s += num12s
        return num1s
    num1s = getNum1s(M)

    ans = pows[N-1]
    for S in range(1, 1<<M):
        num = pows[N-1-numEs[S]]
        if num1s[S] % 2:
            ans -= num
        else:
            ans += num

    print(ans)


solve()
