You are given a tree (a connected undirected graph without cycles) of n vertices. Each of the n - 1 edges of the tree is colored in either black or red.

You are also given an integer k. Consider sequences of k vertices. Let's call a sequence [a_1, a_2, …, a_k] good if it satisfies the following criterion:

  * We will walk a path (possibly visiting same edge/vertex multiple times) on the tree, starting from a_1 and ending at a_k. 
  * Start at a_1, then go to a_2 using the shortest path between a_1 and a_2, then go to a_3 in a similar way, and so on, until you travel the shortest path between a_{k-1} and a_k.
  * If you walked over at least one black edge during this process, then the sequence is good. 

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Consider the tree on the picture. If k=3 then the following sequences are good: [1, 4, 7], [5, 5, 3] and [2, 3, 7]. The following sequences are not good: [1, 4, 6], [5, 5, 5], [3, 7, 3].

There are n^k sequences of vertices, count how many of them are good. Since this number can be quite large, print it modulo 10^9+7.

Input

The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100), the size of the tree and the length of the vertex sequence.

Each of the next n - 1 lines contains three integers u_i, v_i and x_i (1 ≤ u_i, v_i ≤ n, x_i ∈ \{0, 1\}), where u_i and v_i denote the endpoints of the corresponding edge and x_i is the color of this edge (0 denotes red edge and 1 denotes black edge).

Output

Print the number of good sequences modulo 10^9 + 7.

Examples

Input


4 4
1 2 1
2 3 1
3 4 1


Output


252

Input


4 6
1 2 0
1 3 0
1 4 0


Output


0

Input


3 5
1 2 1
2 3 0


Output


210

Note

In the first example, all sequences (4^4) of length 4 except the following are good: 

  * [1, 1, 1, 1]
  * [2, 2, 2, 2]
  * [3, 3, 3, 3]
  * [4, 4, 4, 4] 



In the second example, all edges are red, hence there aren't any good sequences.
class UnionFind:
 
    def __init__(self, n):
        self.follow = [-1]*(n+1)
        self.num_follower = [1]*(n+1)
 
    def root_index_of(self, a):
        r = a
        while self.follow[r] > -1:
                 r = self.follow[r]
        return r
 
    def connected(self, a, b):
        return self.root_index_of(a) == self.root_index_of(b)
 
    def connect(self, a, b):
        ra = self.root_index_of(a)
        rb = self.root_index_of(b)
 
        if ra == rb:
            return
 
        if self.num_follower[ra] < self.num_follower[rb]:
            self.follow[ra] = rb
            self.follow[a] = rb
            self.num_follower[rb] += self.num_follower[ra]
        else:
            self.follow[rb] = ra
            self.follow[b] = ra
            self.num_follower[ra] += self.num_follower[rb]

mod = 10**9+7
n, k = map(int, input().split())

b = 0
r = 0
uf = UnionFind(n)
for _ in range(n-1):
    u, v, color = map(int, input().split())
    if color:
        b += 1
    else:
        uf.connect(u, v)
        r += 1

if b == 0:
    print(0)
    exit()

s = set()
cnt = 0
for i in range(1, n+1):
    root = uf.root_index_of(i)
    num = uf.num_follower[root]
    if root in s:
        continue
    else:
        s.add(root)
        cnt += pow(num, k, mod)

ans = pow(n, k, mod)
ans -= cnt
print(ans%mod)
