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<cstdio>
#include<vector>
#define ll long long

using namespace std;

const int N = 55;
const int M = 3000;
int n,m;
int dep[N];
int fa[N];
vector<int> adj[N];
ll record[N];

ll ppow(ll a,ll b){
	ll res = 1;
	while(b){
		if(b&1) res *= a;
		a = a * a;
		b >>= 1;
	}
	return res;
}

void dfs(int u,int pre){
	fa[u] = pre;
	dep[u] = dep[pre]+1;
	for(int i = 0;i < adj[u].size();i++){
		int v = adj[u][i];
		if(v == pre) continue;
		dfs(v,u);
	}
}

int main(){
	//freopen("in","r",stdin);
	scanf("%d",&n);
	for(int i = 1;i < n;i++){ int u,v;scanf("%d%d",&u,&v);adj[u].push_back(v);adj[v].push_back(u);}
	dfs(1,0);
	scanf("%d",&m);
	for(int i = 0;i < m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		if(dep[u]<dep[v]) swap(u,v);
		while(dep[u]>dep[v]) {record[i] |= (1ll<<u>>1),u = fa[u];}
		while(u != v) {record[i] |= (1ll<<u>>1)|(1ll<<v>>1),u = fa[u],v = fa[v];}
		//record[i] |= (1<<u);
	}	

	ll res = 0;
	for(ll i = 0;i < (1ll<<m);i++){
		ll r = 0;
		int cnt1=0,cnt2=0;
		for(int j = 0;j < m;j++){
			if((i>>j)&1){
				r |= record[j];
				cnt1++;
			}
		}
		for(int j = 0;j < n;j++) if((r>>j)&1) cnt2++;
//		int cnt1 = __builtin_popcount(i);
//		int cnt2 = __builtin_popcount(r);
		res += ((cnt1&1)? -1 : 1) * (1ll << (n-1-cnt2));
	}
	printf("%lld\n", res);
	return 0;
}
