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 int long long
#define pi pair<int,int>
#define ff first
#define ss second
#define boost ios_base::sync_with_stdio(false);cin.tie(nullptr)

const int N = 55;
vector<int>g[N];
int path[N][N];
int cnt = 0;
bool dfs(int u,int p,int d){
    if(u==d)
        return true;
    for(int v:g[u]){
        if(v==p)
            continue;
        if(dfs(v,u,d)){
            cnt |= (1ll<<path[u][v]);
            return true;
        }
    }
    return false;
}
 
int32_t main() { 
    boost;
    int n;
    cin>>n;
    for(int i=0,u,v;i+1<n;i++){
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
        path[u][v] = path[v][u] = i;
    }
    int m;
    cin>>m;
    int a[m];
    for(int i=0;i<m;i++){
        int u,v;
        cin>>u>>v;
        cnt = 0;
        dfs(u,0,v);
        a[i] = cnt;
    }

    int ans = 0;
    for(int i=0;i<(1ll<<m);i++){
        cnt = 0;
        for(int j = 0;j<m;j++){
            if(i&(1ll<<j))
                cnt |= a[j];
        }
        cnt = __builtin_popcountll(cnt);
        int tt = (1ll<<(n-1-cnt));
        cnt = __builtin_popcount(i);
        if(cnt % 2 )
            ans-=tt;
        else
            ans+=tt;
    }
    cout<<ans<<endl;
}