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 rep(i,n) for (long long i = 0; i < (n); ++i)
//g++ -std=c++11 

using namespace std;

long long N, M;
long long A[55];
long long B[55];
vector<long long> tree[55];
long long U[25];
long long V[25];

//えだ、状態
long long dp[55][1048580];

long long cut[55];

//
bool dfs(long long pa, long long cur, long long tar) {
	if(cur == tar) {
		return true;
	}
	for(auto next: tree[cur]) {
		if(next != pa) {
			if(dfs(cur, next, tar)) {
				return true;
			}
		}
	}
	return false;
}

int main(){
	cin >> N;
	rep(i, N-1) {
		cin >> A[i] >> B[i];
		A[i]--;B[i]--;
		tree[A[i]].push_back(B[i]);
		tree[B[i]].push_back(A[i]);
	}
	cin >> M;
	rep(i, M) {
		cin >> U[i] >> V[i];
		U[i]--;V[i]--;
	}

	rep(i, N-1) {
		rep(j, M) {
			long long u = U[j];
			long long v = V[j];

			//uがA側にある && vがB側にある
			if(dfs(B[i], A[i], u) && dfs(A[i], B[i], v)) {
				cut[i] |= (1<<j);
			}
			//uがB側にある && vがA側にある
			if(dfs(B[i], A[i], v) && dfs(A[i], B[i], u)) {
				cut[i] |= (1<<j);
			}
		}
	}

	dp[0][0] = 1;
	rep(i, N-1) {
		rep(j, (1<<M)) {
			//使う
			dp[i+1][j|cut[i]] += dp[i][j];
			//使わない
			dp[i+1][j] += dp[i][j];
		}
	}

	cout << dp[N-1][(1<<M) - 1] << endl;
}