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
// 解答不能.
// betrue12氏.
// https://atcoder.jp/contests/abc152/submissions/9635499
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define repex(i, a, b, c) for(int i = a; i < b; i += c)
#define repx(i, a, b) repex(i, a, b, 1)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)
#define pb push_back

vector<pair<int, int>> edges[50];

int mask[50];
bool dfs(int i, int p, int target, int ei, int b){
    bool ok = (i == target);
    for(auto& e : edges[i]) if(e.first != p) ok |= dfs(e.first, i, target, e.second, b);
    if(ok && ei >= 0) mask[ei] |= 1<<b;
    return ok;
}

int main(){
    int N, M;
    scanf("%d", &N);
    rep(i, N - 1){
        int a, b;
        scanf("%d %d", &a, &b);
        edges[a - 1].pb({b - 1, i});
        edges[b - 1].pb({a - 1, i});
    }
    scanf("%d", &M);
    rep(i, M){
        int a, b;
        scanf("%d %d", &a, &b);
        dfs(a - 1, -1, b - 1, -1, i);
    }
    LL ans = 0;
    rep(b, 1 << M){
        LL res = 1;
        rep(i, N - 1) if(!(b & mask[i])) res *= 2;
        if(__builtin_popcount(b) % 2)    res *= -1;
        ans += res;
    }
    printf("%lld\n", ans);
    return 0;
}