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 pb push_back
using ll = long long;
using namespace std;

using edge = pair<int,int>;
vector<vector<edge>> G;

ll dfs(int v, int p, int g, ll num){
    if (v == g) return num;
    for (auto nv : G[v]) {
        if (nv.first == p) continue;
        ll next_num = num | (1ll<<nv.second);
        ll res = dfs(nv.first, v, g, next_num);
        if (res > 0ll) return res;
    }
    return 0ll;
}

int main() {
    int n; cin >> n;
    G.resize(n);
    for (int i = 0; i < n - 1; i++) {
        int a, b; cin >> a >> b; a--; b--;
        G[a].pb(edge(b,i));
        G[b].pb(edge(a,i));
    }

    int m; cin >> m;
    vector<ll> dist(m);
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b; a--; b--;
        dist[i] = dfs(a, -1, b, 0ll);
    }

    ll ans = 0ll;
    for (int i = 0; i < 1<<m; i++) {
        ll num = 0ll;
        bool check = true;
        for (int j = 0; j < m; j++) if (i & (1<<j)) { num |= dist[j]; check = !check; }
        int k = n - 1 - (int)__builtin_popcountl(num);
        if (check) ans += 1ll<<k;
        else ans -= 1ll<<k;
    }

    cout << ans << endl;
    return 0;
}