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

#define pb push_back
#define fi first
#define SZ(x) (int((x).size()))
#define se second
#define ll long long
#define pq priority_queue
#define MP make_pair
#define pii pair<int,int>
#define mod 998244353
#define debug(x) cerr<<#x<<"="<<x<<'\n'

int n,m;
vector <int> edge[55];
int dep[55],f[55];
pii res[22];
bool vis[55];

void dfs(int u,int lst) {
	f[u]=lst;
	dep[u]=dep[lst]+1;
	for (int i=0;i<SZ(edge[u]);i++) {
		int v=edge[u][i];
		if (v==lst) continue;
		dfs(v,u);
	}
	return;
}

ll ans=0ll;
int cnt;

void paint(int x,int y) {
	if (dep[x]<dep[y]) swap(x,y);
	while (dep[x]>dep[y]) {
		if (!vis[x]) vis[x]=true,cnt++;
		x=f[x];
	}
	while (x!=y) {
		if (!vis[x]) vis[x]=true,cnt++;
		if (!vis[y]) vis[y]=true,cnt++;
		x=f[x],y=f[y];
	}
	return;
}

ll solve(int mask) {
	memset(vis,0,sizeof(vis));
	cnt=0;
	for (int i=0;i<m;i++)
		if ((mask>>i)&1) paint(res[i].fi,res[i].se);
	return 1ll<<(ll)(n-1-cnt);
}

int main(){
	scanf("%d",&n);
	for (int i=0;i<n-1;i++) {
		int u,v;
		scanf("%d%d",&u,&v);
		edge[u].pb(v);
		edge[v].pb(u);
	}
	scanf("%d",&m);
	for (int i=0;i<m;i++) scanf("%d%d",&res[i].fi,&res[i].se);
	dfs(1,0);
	for (int i=0;i<(1<<m);i++) 
		if (__builtin_popcount(i)%2==0) ans+=solve(i);
		else ans-=solve(i);
	printf("%lld",ans);
	return 0;
}
