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. 

<image>

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.par = [int(i) for i in range(n)]
        self.rank = [1] * (n + 1)
        self.rank[0] = 0
    
    def union(self, x, y):
        if not self.same_set(x, y):
            par_x = self.find(x)
            par_y = self.find(y)

            if self.rank[par_x] > self.rank[par_y]:
                self.rank[par_x] += self.rank[par_y]
                self.rank[par_y] = 0
                self.par[par_y] = par_x
            else:
                self.rank[par_y] += self.rank[par_x]
                self.rank[par_x] = 0
                self.par[par_x] = par_y
            self.find(x)
            self.find(y)

    def find(self, x):
        if self.par[x] == x: return x
        self.par[x] = self.find(self.par[x])
        return self.par[x]

    def same_set(self, x, y):
        return self.find(x) == self.find(y)

def fast_pow(x, y):
    if y == 1: return x
    if y % 2 != 0: return ((x % M) * (fast_pow(x, y - 1) % M)) % M
    aux = fast_pow(x, y / 2)
    return ((aux % M) * (aux % M)) % M

M = 1000000007
n, k = map(int, input().split())

max_pos = fast_pow(n, k)

uf = UnionFind(n+1)

for i in range(n - 1):
    u, v, x = map(int, input().split())
    
    if x == 0:
        uf.union(u, v)
nao_bons = 0
for i in range(n+1):
    if uf.rank[i] > 0:
        nao_bons = ((nao_bons % M) + (fast_pow(uf.rank[i], k) % M)) % M
print(((max_pos % M) - (nao_bons % M)) % M)
