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>
using namespace std;
typedef long long ll;
const int N = 52;
const int BIT = (1<<20);
int n, m, d[N];
vector<int> g[N];
ll bit[BIT];
vector<int> v;
void dfs_pre(int fa, int u, int src, int dest, int idx) {
    v.push_back(u);
    if(u == dest) {
        for(int i = 1; i < v.size(); i++) {
            int t = v[i];
            if(d[v[i]] < d[v[i-1]]) t = v[i-1];
            bit[1<<idx] |= (1LL<<t);
        }
    }
    for(int v : g[u]) {
        if(v == fa) continue;
        if(dest==-1) d[v] = d[u]+1;
        dfs_pre(u, v, src, dest, idx);
    }
    v.pop_back();
}
int get_count(ll x) {
    int t = 0;
    while(x) {
        if(x&1) t++;
        x /= 2;
    }
    return t;
}
int main() {
    cin >> n;
    for(int i = 1, u, v; i < n; i++) {
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    cin >> m;
    dfs_pre(-1, 1, 1, -1, 0);
    for(int i = 0, u, v; i < m; i++) {
        cin >> u >> v;
        dfs_pre(-1, u, u, v, i);
    }
    ll ans = (1LL<<(n-1));
    for(int i = 1; i < (1<<m); i++) {
        bit[i] = bit[i&-i] | bit[i-(i&-i)];
        int c = get_count(bit[i]);
        ll res = (1LL<<(n-1-c)) * ((get_count(i) & 1) ? 1 : -1);
        ans -= res;
    }
    cout << ans << '\n';
    return 0;
}