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>
#define LL long long
using namespace std;
inline int read(){
    int x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0',c = getchar();
    return x;
}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
inline void write(int x){if (x > 9) write(x/10); putchar(x%10+'0'); }
const int N = 55;
int n,ex[N],ey[N],G[N][N],fa[N],dpt[N];

int st[N],m,L;
inline void dfs(int x){
	for (int i = 1; i <= n; ++i) if (i!=fa[x]&&G[x][i]>0){
		fa[i] = x; dpt[i] = dpt[x] + 1; dfs(i);
	}
}
inline void Tag(int u,int v,int t){
	while (u != v){
		if (dpt[u] >= dpt[v]) st[G[u][fa[u]]] |= t,u = fa[u];
		else st[G[v][fa[v]]] |= t,v = fa[v];
	}
}
LL dp[1<<20];
inline void DP(int s){
	for (int i = L-1; i >= 0; --i) dp[i|s] += dp[i];
}
int main(){
	int i;
	n = read();
	for (i = 1; i < n; ++i) ex[i] = read(),ey[i] = read(),G[ex[i]][ey[i]] = G[ey[i]][ex[i]] = i;
	dfs(1);
	m = read(),L = 1<<m;
	for (i = 0; i < m; ++i) ex[n] = read(),ey[n] = read(),Tag(ex[n],ey[n],1<<i);
	dp[0] = 1;
	for (i = 1; i < n; ++i) DP(st[i]);
	cout << dp[L-1] << endl;
    return 0;
}