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;
#define ll long long
#define ull unsigned long long
#define mp make_pair
typedef pair<int,int> pii;
typedef priority_queue<int,vector<int>,greater<int> > small_heap;
typedef priority_queue<int> big_heap;
int n,m;
vector<int>v[60];
ll path[60];
void dfs(int x,int fa,ll num){
    path[x]=num;
    for(auto vv:v[x]){
        if(vv==fa)continue;
        dfs(vv,x,num|(1ll<<vv));
    }
}
ll info[60];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n-1;i++){
        int x,y;scanf("%d%d",&x,&y);
        v[x].push_back(y);v[y].push_back(x);
    }
    dfs(1,0,0ll);
    scanf("%d",&m);
    for(int i=0;i<m;i++){
        int x,y;scanf("%d%d",&x,&y);
        info[i]=(path[x]^path[y]);
    }
    ll maxstate=(1ll<<m)-1;
    ll ans=0ll;
    for(int i=0;i<=maxstate;i++)
    {
        ll num=0ll;
        int cnt=0;
        for(int j=0;j<=m-1;j++)
        {
            if(i&(1<<j))
            {
                cnt++;
                num|=info[j];
            }
        }
        int free=n-1-__builtin_popcountll(num);
        ans+=(1ll<<free) * (cnt & 1 ? -1ll : 1ll);//这边奇为减，是因为直接计算合法的总数，2^ n-1  --
    }
    cout<<ans<<endl;
    return 0;
}