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 Graph:
 
    # init function to declare class variables
    def __init__(self, V):
        self.V = V
        self.adj = [[] for i in range(V)]
 
    def DFSUtil(self,temp,v, visited):
 
        # Mark the current vertex as visited
        visited[v] = True
 
        # Store the vertex to list
        
        temp.append(v)
 
        # Repeat for all vertices adjacent
        # to this vertex v
        for i in self.adj[v]:
            if visited[i] == False:
 
                # Update the list
                temp = self.DFSUtil(temp, i, visited)
        return temp
 
    # method to add an undirected edge
    def addEdge(self, v, w):
        self.adj[v].append(w)
        self.adj[w].append(v)
 
    # Method to retrieve connected components
    # in an undirected graph
    def connectedComponents(self):
        visited = []
        cc = []
        for i in range(self.V):
            visited.append(False)
        for v in range(self.V):
            if visited[v] == False:
                temp = []
                cc.append(len(self.DFSUtil(temp, v, visited)))
        return cc
n,k = map(int,input().split())
g = Graph(n)
for i in range(n-1):
  u,v,x = map(int,input().split())
  if x == 0:
    g.addEdge(u-1,v-1)
L = g.connectedComponents() 
def power(x, y, p) : 
    res = 1     # Initialize result 
  
    # Update x if it is more 
    # than or equal to p 
    x = x % p  
      
    if (x == 0) : 
        return 0
  
    while (y > 0) : 
          
        # If y is odd, multiply 
        # x with result 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
  
        # y must be even now 
        y = y >> 1      # y = y/2 
        x = (x * x) % p 
          
    return res 
p = 1e9 + 7
alpha = power(n,k,p)
for i in L:
  alpha -= power(i,k,p)
  alpha = alpha%p
print(int(alpha%p))