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>
#define va first
#define vb second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

vector<pii> G[51];
int par[51],dep[51],ord[51];

void dfs(int s)
{
	for(pii p:G[s]){
		int e = p.va;
		if(e==par[s]) continue;
		par[e] = s;
		ord[e] = p.vb;
		dep[e] = dep[s]+1;
		dfs(e);
	}
}

int main()
{
    ios_base::sync_with_stdio(0),cin.tie(0);
    int N,M;
    cin >> N;

    ll K[N];
    K[0] = 1;
    for(int i=1; i<N; i++){
    	K[i] = K[i-1]*2;
    }

    int u,v;
    for(int i=1; i<N; i++){
    	cin >> u >> v;
    	G[u].emplace_back(v,i);
    	G[v].emplace_back(u,i);
    }

    dfs(1);

    cin >> M;
    bitset<51> B[1<<M];

    for(int i=0; i<M; i++){
    	cin >> u >> v;

    	if(dep[v]>dep[u]) swap(u,v);


    	while(dep[u]>dep[v]){
    		B[1<<i][ord[u]] = 1;
    		u = par[u];
    	}
    	while(u!=v){
    		B[1<<i][ord[u]] = 1;
    		u = par[u];
    		B[1<<i][ord[v]] = 1;
    		v = par[v];    		
    	}
    }

    ll ans = K[N-1];

    for(int i=1; i<(1<<M); i++){

    	B[i] = B[i-(i&-i)]|B[i&-i];

    	int cnt = N-1;
    	for(int j=1; j<N; j++){
    		cnt -= B[i][j];
    	}

    	if(__builtin_popcount(i)%2) ans -= K[cnt];
    	else ans += K[cnt];
    }

    cout << ans;
}