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.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
char *fs, *ft, buf[1 << 20];
inline long long read() {
  long long x = 0, f = 1;
  char ch =
      (fs == ft && (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
          ? 0
          : *fs++;
  ;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
             ? 0
             : *fs++;
    ;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
             ? 0
             : *fs++;
    ;
  }
  return x * f;
}
using namespace std;
const long long N = 1e6 + 1000;
const long long inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const double eps = 1e-6;
const double PI = acos(-1);
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow1(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow2(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow3(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow4(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow5(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow6(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow7(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow8(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow9(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow10(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow11(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow12(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow13(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow14(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow15(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow16(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow17(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow18(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow19(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow20(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow21(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow22(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow23(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow24(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow25(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow26(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long qpow27(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
vector<long long> e[N], tmp;
long long siz[N], n, num[N];
void dfs(long long u, long long fa) {
  siz[u] = 1;
  for (auto v : e[u]) {
    if (v == fa) continue;
    dfs(v, u);
    siz[u] += siz[v];
    tmp.push_back((n - siz[v]) * siz[v]);
  }
}
signed main() {
  long long T;
  cin >> T;
  while (T--) {
    cin >> n;
    for (long long i = 1; i <= n; i++) e[i].clear();
    for (long long i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    tmp.clear();
    dfs(1, 1);
    long long m;
    cin >> m;
    for (long long i = 1; i <= m; i++) cin >> num[i];
    sort(num + 1, num + 1 + m, [](long long x, long long y) { return x > y; });
    long long ans = 0;
    sort(tmp.begin(), tmp.end(),
         [](long long x, long long y) { return x > y; });
    if (m >= n - 1) {
      long long cnt = m - (n - 1);
      long long i = 1;
      while (cnt--) {
        num[i + 1] *= num[i];
        num[i + 1] %= mod;
        i++;
      }
      long long j = 0;
      for (i; i <= m; i++) ans = (ans + num[i] * tmp[j++]) % mod;
    } else {
      for (long long i = m + 1; i <= n - 1; i++) num[i] = 1;
      long long j = 0, i = 1;
      for (i; i <= n - 1; i++) {
        ans = (ans + num[i] * tmp[j++]) % mod;
      }
    }
    cout << ans << '\n';
  }
}
