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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

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

void init(int x, int d) {
    depth[x] = d;
    for (auto &i : v[x]) {
        if (i.first != p[x]) {
            p[i.first] = x;
            init(i.first, d + 1);
        }
    }
}

void lca(int x, int y, int z) {
    if (depth[x] > depth[y]) swap(x, y);
    while (depth[y] > depth[x]) {
        for (auto &i : v[y]) {
            if (i.first == p[y]) {
                c[i.second] |= 1 << z;
            }
        }
        y = p[y];
    }
    while (y != x) {
        for (auto &i : v[x]) {
            if (i.first == p[x]) {
                c[i.second] |= 1 << z;
            }
        }
        x = p[x];
        for (auto &i : v[y]) {
            if (i.first == p[y]) {
                c[i.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][0] = 1;
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < 1 << m; j++) {
            dp[i + 1][j | c[i]] += dp[i][j];
            dp[i + 1][j] += dp[i][j];
        }
    }

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