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
#include<cstdio>
#include<utility>
#include<vector>
using namespace std;
typedef pair<int, int> P;

int n, m;
int c[49], depth[50], p[50], ind[50];
long dp[1 << 20];
vector<P> v[50];

void init(int x, int d) {
    depth[x] = d;
    for (int i = 0; i < v[x].size(); i++) {
        if (v[x][i].first != p[x]) {
            p[v[x][i].first] = x;
            init(v[x][i].first, d + 1);
        }
        else ind[x] = i;
    }
}

void lca(int x, int y, int z) {
    if (depth[x] > depth[y]) swap(x, y);
    while (depth[y] > depth[x]) {
        c[v[y][ind[y]].second] |= 1 << z;
        y = p[y];
    }
    while (y != x) {
        c[v[x][ind[x]].second] |= 1 << z;
        x = p[x];
        c[v[y][ind[y]].second] |= 1 << z;
        y = p[y];
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        v[a - 1].emplace_back(b - 1, i);
        v[b - 1].emplace_back(a - 1, i);
    }
    scanf("%d", &m);
    init(0, 0);
    for (int i = 0; i < m; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        lca(u - 1, v - 1, i);
    }
    
    dp[0] = 1;
    for (int i = 0; i < n - 1; i++) {
        for (int j = (1 << m) - 1; j >= 0; j--) dp[j | c[i]] += dp[j];
    }

    printf("%ld\n", dp[(1 << m) - 1]);
    return 0;
}
