Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=55;
int n,m,f[N],U[N],V[N],vis[N][N],tot,head[N],nex[N<<1],to[N<<1];
ll dp[N][1<<20];
bool vv[N];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
void dfs(int u,int p)
{
    vv[u]=true;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p)continue;
        dfs(v,u);
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        scanf("%d%d",&U[i],&V[i]);
        add(U[i],V[i]);add(V[i],U[i]);
    }
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        vis[u][v]=vis[v][u]=i;
    }
    for(int i=1;i<n;i++)
    {
        memset(vv,false,sizeof(vv));
        dfs(U[i],V[i]);
        for(int j=1;j<=n;j++)
            for(int k=1;k<=n;k++)
            if(vv[j]!=vv[k]&&vis[j][k])
        {
            f[i]|=(1<<vis[j][k]-1);
        }
    }
    dp[0][0]=1;
    int up=1<<m;
    for(int i=0;i<n-1;i++)
        for(int j=0;j<up;j++)
        if(dp[i][j])
        {
            dp[i+1][j]+=dp[i][j];
            dp[i+1][j|f[i+1]]+=dp[i][j];
        }
    printf("%lld\n",dp[n-1][up-1]);
}