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 REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define REV(i,n) for(ll (i) = (n) - 1;(i) >= 0;--i)
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {REP(WW,n)cerr << v[WW] << ' ';cerr << endl << endl;}
#define SHOW2d(v,WW,HH) {REP(W_,WW){REP(H_,HH)cerr << v[W_][H_] << ' ';cerr << endl;}cerr << endl;}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(20)
#define INF 1000000000
#define LLINF 1000000000000000000LL
#define MOD 998244353

typedef long long ll;
typedef pair<ll,ll> P;

int mask[55];
ll dp[55][1<<20];

vector<vector<P>> v(55);

bool dfs(int now, int goal, int pre, int query_id){
	if(now == goal)return true;
	bool flag = false;
	REP(i, v[now].size()){
		int aite = v[now][i].FI;
		int edge_id = v[now][i].SE;
		if(aite == pre)continue;
		if(dfs(aite, goal, now, query_id)){
			mask[edge_id] |= (1 << query_id);
			flag = true;
		}
	}
	return flag;
}

int main(){
	cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);

	ll n, m;cin >> n;
	REP(i, n-1){
		ll a, b;cin >> a >> b;
		a--;b--;
		v[a].EB(b, i);
		v[b].EB(a, i);
	}
	cin >> m;
	REP(i, m){
		ll a, b;cin >> a >> b;
		a--;b--;
		dfs(a, b, -1, i);
	}

	dp[0][0] = 1;
	REP(i, n-1){
		REP(j, (1<<m))if(dp[i][j]){
			dp[i+1][j | mask[i]] += dp[i][j];
			dp[i+1][j] += dp[i][j];
		}
	}

	cout << dp[n-1][(1<<m)-1] << endl;

	return 0;
}