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 <bits/stdc++.h>
#define rep(i,n) for(int i=(0);i<(n);i++)

using namespace std;

typedef long long ll;

using edge = pair<int, int>;

const int max_n = 55;
int n, m;
vector<edge> g[max_n];
vector<int> us, vs;

bool rec(int v, int p, int target, ll &path){
	if(v == target){
		path = 0;
		return true;
	}

	for(auto e : g[v]){
		if(e.first == p) continue;
		if(rec(e.first, v, target, path)){
			path |= (1ll << e.second);
			return true;
		}
	}

	return false;
}

ll solve(){
	vector<ll> paths(m, 0);
	rep(i, m) rec(us[i], -1, vs[i], paths[i]);

	ll res = 0;
	for(ll bit = 0; bit < (1 << m); bit++){
		ll val = 0;
		rep(i, m) if((bit >> i) & 1ll) val |= paths[i];
		ll remnum = n - 1 - __builtin_popcountl(val);
		if(__builtin_popcountl(bit) % 2 == 0) res += (1ll << remnum);
		else res -= (1ll << remnum);
	}

	return res;
}

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

	cin >> n;

	rep(i, n-1){
		int a, b;
		cin >> a >> b;
		a--; b--;
		g[a].push_back({b, i});
		g[b].push_back({a, i});
	}

	cin >> m;
	us.resize(m);
	vs.resize(m);
	rep(i, m){
		cin >> us[i] >> vs[i];
		us[i]--; vs[i]--;
	}

	cout << solve() << endl;
}
