Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, a, b) for (int i = (a); i < (b); i++)
//#include "library/debug.cpp"

vector<vector<ll>> g(55);
ll n, m;
map<pair<ll, ll>, ll> mp;

ll dfs(ll a, ll b, ll p = -1) {
    ll res = 0;
    for (auto u : g[a])
        if (u != p) {
            if (u == b) {
                res = 1LL << mp[{a, b}];
                return res;
            }
            res = dfs(u, b, a);
            if (res) {
                res |= 1LL << mp[{a, u}];
                break;
            }
        }
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> n;
    rep(i, 1, n) {
        ll a, b;
        cin >> a >> b;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
        mp[{a, b}] = mp[{b, a}] = i - 1;
    }
    cin >> m;
    vector<ll> y(m);
    rep(i, 0, m) {
        ll u, v;
        cin >> u >> v;
        y[i] = dfs(u, v);
    }
    ll ans = 1LL << (n - 1);
    for (ll b = 1; b < (1 << m); b++) {
        ll d = 0;
        ll c = 1;
        rep(i, 0, m) if ((b >> i) & 1) {
            c *= -1;
            d |= y[i];
        }
        ans += c * (1LL << (n - 1 - __builtin_popcountll(d)));
    }
    cout << ans << endl;
    return 0;
}