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: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define pb push_back
#define ll long long
#define maxn 200005
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define ms(i) memset(i,0,sizeof(i));
#define ms1(i) memset(i,-1,sizeof(i));
#define F first
#define S second
const int mod = 1e9+7;
vector<int>g[55];
int dep[55];
int pa[55];
ll sol[55];
void dfs(int now, int pre){
    for(auto i:g[now]){
        if(i != pre){
            pa[i] = now;
            dep[i] = dep[now] + 1;
            dfs(i,now);
        }
    }
}
ll solve(int x,int y){
    if(dep[x] < dep[y]){
        swap(x, y);
    }
    ll ret = 0;
    while(dep[x] > dep[y]){
        ret |= (1LL<<x);
        x = pa[x];
    }
    while(x != y){
        ret |= (1LL<<x);
        x = pa[x];
        ret |= (1LL<<y);
        y = pa[y];
    }
    return ret;
}
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    while(cin >> n){
        f(n-1){
            int add1, add2;
            cin >> add1 >> add2;
            add1--;
            add2--;
            g[add1].pb(add2);
            g[add2].pb(add1);
        }
        dfs(0,0);
        int m;
        cin >> m;
        f(m){
            int add1, add2;
            cin >> add1 >> add2;
            add1--, add2--;
            sol[i] = solve(add1, add2);
        }
        ll ans = 0;
        for(int i = 0 ; i < (1 << m) ; i++){
            ll mask = 0;
            for(int j = 0 ; j < m ; j ++){
                if(i&(1<<j)){
                    mask |= sol[j];
                }
            }
            int cnt = n - 1 - __builtin_popcountll(mask);
            if(__builtin_popcount(i)&1){
                ans -= (1LL<<cnt);
            }
            else{
                ans += (1LL<<cnt);
            }
        }
        cout << ans << '\n';
    }
}