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)

const int isDebugMode = 1;
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<long long, long long> P;
struct edge{long long to,cost;};
const int inf = 1 << 27;
const long long INF = 1LL << 60;
const int COMBMAX = 1001001;
const long long MOD = 1000000007;
const long long dy[] = {-1, 0, 0, 1};
const long long dx[] = {0, -1, 1, 0};
string abc = "abcdefghijklmnopqrstuvwxyz";
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define eachdo(v, e) for (const auto &e : (v))
#define all(v) (v).begin(), (v).end()
#define lower_index(v, e) (long long)distance((v).begin(), lower_bound((v).begin(), (v).end(), e))
#define upper_index(v, e) (long long)distance((v).begin(), upper_bound((v).begin(), (v).end(), e))
long long mpow(long long a, long long n, long long mod = MOD){long long res = 1; while(n > 0){if(n & 1)res = res * a % mod; a = a * a % mod; n >>= 1;} return res;}
void pt(){cout << endl; return;}
template<class Head> void pt(Head&& head){cout << head; pt(); return;}
template<class Head, class... Tail> void pt(Head&& head, Tail&&... tail){cout << head << " "; pt(forward<Tail>(tail)...);}
void dpt(){if(!isDebugMode) return; cout << endl; return;}
template<class Head> void dpt(Head&& head){if(!isDebugMode) return; cout << head; pt(); return;}
template<class Head, class... Tail> void dpt(Head&& head, Tail&&... tail){if(!isDebugMode) return; cout << head << " "; pt(forward<Tail>(tail)...);}
template<class T> void debug(T v){if(!isDebugMode) return; rep(i, v.size()) cout << v[i] << " " ; cout << endl;}
template<class T> void debug2(T v){if(!isDebugMode) return; rep(i, v.size()){rep(j, v[i].size()) cout << v[i][j] << " " ; cout << endl;}}
template<class T1, class T2> inline bool chmin(T1 &a, T2 b){if(a > b){a = b; return true;} return false;}
template<class T1, class T2> inline bool chmax(T1 &a, T2 b){if(a < b){a = b; return true;} return false;}
template<class T1, class T2> long long recgcd(T1 a, T2 b){return a % b ? recgcd(b , a % b) : b;}
bool valid(long long H, long long W, long long h, long long w) { return 0 <= h && h < H && 0 <= w && w < W; }

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    ll N; cin >> N;
    vector<ll> a(N - 1), b(N - 1), c(N - 1);
    rep(i, N - 1){
        cin >> a[i] >> b[i];
        a[i]--;
        b[i]--;
        c[i] = 1LL << i;
    }
    ll M; cin >> M;
    vector<ll> u(M), v(M);
    rep(i, M){
        cin >> u[i] >> v[i];
        u[i]--;
        v[i]--;
    }
    vector<vector<ll>> dist(N, vector<ll>(N, INF));
    rep(i, N) dist[i][i] = 0;
    rep(i, N - 1){
        dist[a[i]][b[i]] = c[i];
        dist[b[i]][a[i]] = c[i];
    }
    rep(k, N)rep(i, N)rep(j, N){
        chmin(dist[i][j], dist[i][k] + dist[k][j]);
    }
    ll ans = 0;
    for(ll S = 1; S < (1 << M); S++){
        ll path = 0;
        for(ll i = 0; i < M; i++){
            if(S >> i & 1){
                path |= dist[u[i]][v[i]];
            }
        }
        ll on = __builtin_popcountll(path);
        if (__builtin_popcountll(S) % 2){
            ans += 1LL << (N - 1 - on);
        }else{
            ans -= 1LL << (N - 1 - on);
        }
    }
    pt((1LL << N - 1) - ans);
    return 0;  
}