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:
#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

ll dp[1 << 20];

ll V[51];
vector<pair<int, int>> E[51];
void dfs(int p, ll v, int mae) {
	V[p] = v;
	for (auto pr : E[p]) {
		int to = pr.first;
		if (mae != to) {
			v ^= 1ll << pr.second;
			dfs(to, v, p);
			v ^= 1ll << pr.second;
		}
	}
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	int N;
	cin >> N;
	rep(i, N - 1) {
		int a, b;
		cin >> a >> b;
		E[a].pb(mp(b, i));
		E[b].pb(mp(a, i));
	}

	dfs(1, 0, 0);

	int M;
	cin >> M;
	int list[51] = {};
	rep(i, M) {
		int u, v;
		cin >> u >> v;

		rep(j, N - 1) list[j] ^= (V[u] >> j & 1ll) << i;
		rep(j, N - 1) list[j] ^= (V[v] >> j & 1ll) << i;
	}

	dp[0] = 1;
	rep(i, N - 1) {
		for (int j = (1 << M) - 1; j >= 0; j--) {
			dp[j | list[i]] += dp[j];
		}
	}

	co(dp[(1 << M) - 1]);


	Would you please return 0;
}