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;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
const ll MOD=1e9+7;
const ll MOD2=998244353;
const ll N=55;
const ld pi=3.14159265359;
const ll M=(1LL<<20)+5;
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define setp setprecision
#define lwb lower_bound
#define SZ(a) (ll)a.size()

ll dp[N][M],vis[N],x,y,n,m,a[N],b[N],ans;
ll bit[N];
vector<ll> v[N];
vector<pll> e;

void fil(ll nd,ll p,ll c){
	vis[nd]=c;
	for(auto i:v[nd]){
		if(i!=p)fil(i,nd,c);
	}
}

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);
	cin>>n;
	REP(i,n-1){
		cin>>x>>y;
		e.pb(mp(x,y));
		v[x].pb(y);v[y].pb(x);
	}
	cin>>m;
	REP(i,m)cin>>a[i]>>b[i];
	REP(i,n-1){
		fil(e[i].X,e[i].Y,0);fil(e[i].Y,e[i].X,1);
		REP(j,m)if(vis[a[j]]!=vis[b[j]])bit[j]|=(1LL<<i);
	}
	REP(i,(1LL<<m)){
		ll c=1,base=0,ti=0;
		REP(j,m)if(i&(1LL<<j))base|=(bit[j]),c*=-1;
		REP(j,n-1)if(!(base&(1LL<<j)))ti++;
		ans+=(c*(1LL<<ti));
	}
	cout<<ans<<"\n";
	return 0;
}

