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
#pragma GCC optimize("trapv")
//#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//using namespace __gnu_pbds;
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using vi = vector<ll>;
using pi = pair<ll, ll>;
using vpi = vector<pi>;
const int maxn = 2e5 + 55;
const int mod = 1e9 + 7;
ll n, m, col[60], dp[2][1<<20];
vector<vpi> g;
bool dfs(int v, int p, int t, int c) {
	if(v==t) return true;
	for(auto i : g[v]) {
		if(i.first==p) continue;
		if(dfs(i.first, v, t, c)) {
			col[i.second]|=c;
			return true;
		}
	}
	return false;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n;
	g.resize(n+1);
	for(int f, t, i = 1; i <n ; i++) {
		cin >> f >> t;
		g[f].pb({t, i-1});
		g[t].pb({f, i-1});
	}
	cin >> m;
	for(int f, t, i = 0; i < m; i++) {
		cin >> f >> t;
		dfs(f, f, t, 1<<i);
	}
	memset(dp, 0, sizeof dp);
	dp[0][0] = 1;
	for(int it = 0; it < n-1; it++) {
		int i = it&1;
		memset(dp[i^1], 0, sizeof dp[i^1]);
		for(int c = 1<<m; c--;) {
			dp[i^1][c|col[it]] += dp[i][c];
			dp[i^1][c] += dp[i][c];
		}
	}
	cout << dp[(n-1)&1][(1<<m)-1];

}
