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>
using namespace std;
#define ll long long
#define pb push_back
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define pll pair<ll,ll>
#define sqr(x) (x)*(x)
#define mod 1000000007

ll n,m,tim,dp[52][1048580];
vector<pll> edge,rest;
vector<ll> adj[52],effect[52];
bool vis[52];

void dfs(ll u, ll p){
	vis[u]=1;
	for(auto v:adj[u]){
		if(v!=p) dfs(v,u);
	}
}

ll call(ll pos, ll mask){
	if((pos+1)==n) return mask==((1ll<<m)-1);
	ll &ret=dp[pos][mask];
	if(ret!=-1) return ret;
	ret=0;
	ret+=call(pos+1,mask);
	ll nmask=mask;
	for(int i=0;i<(int)effect[pos].size();i++){
		ll idx=effect[pos][i];
		nmask|=(1ll<<idx);
	}
	ret+=call(pos+1,nmask);
	return ret;
}

int main()
{
    scanf("%lld",&n);
    for(int i=0;i<n-1;i++){
		ll u,v;
		scanf("%lld %lld",&u,&v);
		adj[u].pb(v);
		adj[v].pb(u);
		edge.pb({u,v});
	}
	scanf("%lld",&m);
	for(int i=0;i<m;i++){
		ll u,v;
		scanf("%lld %lld",&u,&v);
		rest.pb({u,v});
	}
	for(int i=0;i<n-1;i++){
		memset(vis,0,sizeof vis);
		dfs(edge[i].ff,edge[i].ss);
		for(int j=0;j<m;j++){
			if(vis[rest[j].ff]!=vis[rest[j].ss]) effect[i].pb(j);
		}
	}
	memset(dp,-1,sizeof dp);
	ll ans=call(0,0);
	printf("%lld\n",ans);
	return 0;
}

