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];
long long U[25];
long long V[25];

//条件i に違反するために白でならなければならない場所
long long state[55];

void dfs(long long pa, long long cur, long long idx, long long s) {
	if(V[idx] == cur) {
		state[idx] = s;
		return;
	}

	rep(i, N - 1) {
		if(A[i] == cur) {
			if(B[i] != pa)
				dfs(cur, B[i], idx, s | (1LL<<i));
		} else if (B[i] == cur) {
			if(A[i] != pa)
				dfs(cur, A[i], idx, s | (1LL<<i));
		}
	}
}


int main(){
	cin >> N;
	rep(i, N - 1) {
		cin >> A[i] >> B[i];
	}
	cin >> M;
	rep(i, M) {
		cin >> U[i] >> V[i];
		dfs(-1, U[i], i, 0);
	}
	long long ans = 0;

	rep(i, (1<<M)) {
		long long white = 0;
		long long zyokennum = 0;
		rep(j, M) {
			//かつ　j番目の条件を満たさない
			if((i>>j)&1) {
				white = white | state[j];
				zyokennum++;
			}
		}

		// white が 白だと 条件が満たされない
		long long whitenum = __builtin_popcountll(white);


		if(zyokennum%2 == 0) {
			ans += pow(2, N - whitenum - 1);
		} else {
			ans -= pow(2, N - whitenum - 1);
		}
	}
	cout << ans << endl;

}

