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;
const ll N=55;
vector<ll> adj[N];
ll mask[N],par[N],dp[1ll<<20][N];
set<ll> st[25];
void dfs(ll src,ll pr){
    par[src]=pr;
    for(auto &v:adj[src])if(v!=pr)dfs(v,src);
}
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n;cin>>n;
    for(ll i=0,u,v;i<n-1;i++)cin>>u>>v,adj[u].push_back(v),adj[v].push_back(u);
    dfs(1,0);
    ll m;cin>>m;
    for(ll i=0;i<m;i++){
        ll a,b;cin>>a>>b;
        while(a!=1)st[i].insert(a),a=par[a];
        while(b!=1){
            if(st[i].count(b))st[i].erase(b);
            else st[i].insert(b);
            b=par[b];
        }
    }

    for(ll i=0;i<(1ll<<m);i++)for(ll j=0;j<=n+1;j++)dp[i][j]=0;
    dp[0][1]=1;
    for(ll i=0;i<=n;i++)for(ll l=0;l<m;l++)if(st[l].count(i))mask[i]=(mask[i]|(1ll<<l));

    for(ll i=0;i<(1ll<<m);i++){
        for(ll j=2;j<=n;j++){
            dp[i][j]+=dp[i][j-1];
            ll k=i|mask[j];
            dp[k][j]+=dp[i][j-1];
        }
    }

    cout<< dp[(1ll<<m)-1][n];
}

