1401_D. Maximum Distributed Tree
You are given a tree that consists of n nodes. You should label each of its n-1 edges with an integer in such way that satisfies the following conditions: 

  * each integer must be greater than 0; 
  * the product of all n-1 numbers should be equal to k; 
  * the number of 1-s among all n-1 integers must be minimum possible. 



Let's define f(u,v) as the sum of the numbers on the simple path from node u to node v. Also, let ∑_{i=1}^{n-1} ∑_{j=i+1}^n f(i,j) be a distribution index of the tree.

Find the maximum possible distribution index you can get. Since answer can be too large, print it modulo 10^9 + 7.

In this problem, since the number k can be large, the result of the prime factorization of k is given instead.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes in the tree.

Each of the next n-1 lines describes an edge: the i-th line contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i) — indices of vertices connected by the i-th edge.

Next line contains a single integer m (1 ≤ m ≤ 6 ⋅ 10^4) — the number of prime factors of k.

Next line contains m prime numbers p_1, p_2, …, p_m (2 ≤ p_i < 6 ⋅ 10^4) such that k = p_1 ⋅ p_2 ⋅ … ⋅ p_m.

It is guaranteed that the sum of n over all test cases doesn't exceed 10^5, the sum of m over all test cases doesn't exceed 6 ⋅ 10^4, and the given edges for each test cases form a tree.

Output

Print the maximum distribution index you can get. Since answer can be too large, print it modulo 10^9+7.

Example

Input


3
4
1 2
2 3
3 4
2
2 2
4
3 4
1 3
3 2
2
3 2
7
6 1
2 3
4 6
7 3
5 1
3 6
4
7 5 13 3


Output


17
18
286

Note

In the first test case, one of the optimal ways is on the following image:

<image>

In this case, f(1,2)=1, f(1,3)=3, f(1,4)=5, f(2,3)=2, f(2,4)=4, f(3,4)=2, so the sum of these 6 numbers is 17.

In the second test case, one of the optimal ways is on the following image:

<image>

In this case, f(1,2)=3, f(1,3)=1, f(1,4)=4, f(2,3)=2, f(2,4)=5, f(3,4)=3, so the sum of these 6 numbers is 18.
```python
import sys
input=sys.stdin.readline
for _ in range(int(input())):
    n=int(input())
    dic={}
    dico={}
    for i in range(1,n+1):
        dic[i]=[]
        dico[i]=[]
    for i in range(n-1):
        x,y=map(int,input().split())
        dic[x].append(y)
        dic[y].append(x)
        dico[x].append(y)
        dico[y].append(x)
    dis={}
    dis[1]=0
    level=1
    se={1}
    ne={1}
    while(ne):
        tem=set({})
        for i in ne:
            xx=dic[i]
            for j in xx:
                if(not(j in se)):
                    se.add(j)
                    dis[j]=level
                    tem.add(j)
        ne=tem.copy()
        level+=1
    for i in dic:
        dic[i].sort(key=lambda x:dis[x])
    people={}
    st=1
    se={1}
    people[1]=1
    while(dic[st]):
        ne=dic[st].pop()
        if(not(ne in se)):
            people[ne]=1
            se.add(ne)
        else:
            people[ne]+=people[st]
        st=ne
    size=[]
    se={1}
    ne={1}
    while(ne):
        tem=set({})
        for i in ne:
            xx=dico[i]
            for j in xx:
                if(not(j in se)):
                    se.add(j)
                    size.append(people[j]*(n-people[j]))
                    tem.add(j)
        ne=tem.copy()
    size.sort(reverse=True)
    m=int(input())
    prime=list(map(int,input().split()))
    prime.sort(reverse=True)
    rem=1000000007
    ans=0
    if(m<=n-1):
        for i in range(m):
            ans=(ans+(prime[i]*size[i])%rem)%rem
        for i in range(m,n-1):
            ans=(ans%rem+size[i]%rem)%rem
    else:
        ext=m-(n-1)
        for i in range(ext):
            prime[i+1]=(prime[i+1]*prime[i])%rem
        for i in range(n-1):
            ans=(ans+(prime[i+ext]*size[i])%rem)%rem
    print(ans)
            
```