Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

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

Sample Output 4

62

The tree in this input is shown below:

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
#include <iomanip>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<long double> VD;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VD> VVD;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }
#define INF 1LL<<60

struct Edge {
    int to, id;

    Edge(int to, int id): to(to), id(id) {}
};

vector<Edge> g[55];
vector<int> es;

bool dfs(int cur, int t, int p = -1) {
    if(cur == t) return true;
    for(Edge e : g[cur]) {
        if(e.to == p) continue;
        if(dfs(e.to, t, cur)) {
            es.push_back(e.id);
            return true;
        }
    }
    return false;
}


int main() {
    int N;
    cin >> N;
    REP(i, N-1) {
        int a, b;
        cin >> a >> b;
        a--; b--;
        g[a].emplace_back(b, i);
        g[b].emplace_back(a, i);
    }

    int M;
    cin >> M;

    vector<ll> eset(M);
    REP(i, M) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        es = vector<int>();
        dfs(u, v);
        for(int e: es) {
            eset[i] |= 1ll<<e;
        }
    }

    ll ans = 0;
    REP(bit, 1<<M) {
        ll mask = 0;
        REP(j, M)
            if(bit >> j & 1) mask |= eset[j];

        int bit_cnt = __builtin_popcountll(mask);
        ll now = 1ll<<(N - 1 - bit_cnt);
        if(__builtin_popcountll(bit)%2 == 0)
            ans += now;
        else
            ans -= now;
    }
    
    printf("%lld\n", ans);
    return 0;
}