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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
using namespace std;

typedef long long ll;
typedef double db;
typedef long double ldb;

const int inf=1e9+10;
const int maxn=55;

ll dp[maxn];
int xx[maxn],n,m;
vector <pair<int,int> > g[maxn];

bool dfs(int x,int fa,int ed,int id){
	if(x==ed) return true; 
	for(int i=0;i<g[x].size();i++){
		int to=g[x][i].fst;
		if(to!=fa) if(dfs(to,x,ed,id)){
			xx[g[x][i].snd]|=1<<id;
			return true;
		}
	}
	return false;
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int u,v; 
		scanf("%d%d",&u,&v);
		u--; v--;
		g[u].pb(mp(v,i));
		g[v].pb(mp(u,i));
	}
	scanf("%d",&m);
	for(int i=0;i<m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		u--; v--;
		dfs(u,-1,v,i);
	}
	dp[0]=1ll;
	for(int i=0;i<n-1;i++) 
		for(int mask=(1<<m)-1;mask>=0;mask--)
			dp[mask|xx[i]]+=dp[mask];
	printf("%lld\n",dp[(1<<m)-1]);
	return 0;
}