Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n) for(int (i) = 0; (i) < (n); ++(i))
using namespace std;
using ll=long long;

struct Edge {
	int to, id;
	Edge(int to, int id) :to(to), id(id) {}
};

vector<Edge> esets[55];

vector<int> es;
int dfs(int st, int to, int prev = -1) {
	if (st == to) return true;
	for(Edge e : esets[st]) {
		if (prev == e.to) continue;
		if ( dfs(e.to, to, st) == true) {
			es.push_back(e.id);
			return true;
		}
	}
	return false;
}


int main(void) {
	int n,m;
	cin >> n;
	rep(i, n-1) {
		int a, b;
		cin >> a >> b;
		--a;
		--b;
		esets[a].emplace_back(b, i);
		esets[b].emplace_back(a, i);
	}
	
	cin >> m;
	vector<ll> combination(m);	// 各条件を白線が有効な場合にbitを立てている
	rep(i, m) {
		int a, b;
		cin >> a >> b;
		es.clear();	// es = vector<int>()
		--a;
		--b;
		dfs(a, b);
		for(int id : es) {
			combination[i] |= 1ll<<id;
		}
	}
	
	ll answer = 0;
	int counter;
	rep(i, 1<<m) {
		ll mask = 0;
		rep(j,m) {
			if (((i>>j)&1)==1) mask |= combination[j];
		}
		int counter = __builtin_popcountll(mask);
		ll now = 1ll << (n-1-counter);
		if (__builtin_popcountll(i)%2==0) answer += now;
		else answer -= now;
		
	}
	
	cout << answer << endl;
	
	return 0;
	
}