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.
def power(x,y,n):
    ans=1
    while y>0 :
        if y&1==1 :
            ans=ans*x%n
        x=x*x%n
        y>>=1
    return(ans)



class disjoint_set:
    def __init__(self,num):

        self.array=[0]+[[i,0,1] for i in range(1,num+1)]  # our set
    
    def union(self,a,b):
        
        ''' a and b are the vertices that we want to connect them to each other '''
        
        a=self.find(a) # finds the ancestor of a
        b=self.find(b) # finds the ancestor of b
    
        if a==b:
            # if the ancestors of a and b are the same , do nothing
            return False
    

        # connecting two graph components to each other

        if self.array[a][1] > self.array[b][1] :
            self.array[b][0]=a
            self.array[a][2]+=self.array[b][2]
        
        elif self.array[a][1] < self.array[b][1] :
            self.array[a][0]=b
            self.array[b][2]+=self.array[a][2]

        else:
            self.array[a][0]=b
            self.array[b][1]+=1
            self.array[b][2]+=self.array[a][2]
        return True
            
    def find(self,a):
        p=[]
        while self.array[a][0]!=a :
            p.append(a)
            a=self.array[a][0]
        for i in p:
            self.array[i][0]=a
        return a

###### start #######

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

graph=disjoint_set(n)
for _ in range(n-1):
    a,b,c=map(int,input().split())
    if not c :
        graph.union(a,b)

s=set()
for i in range(1,n+1):
    s.add(graph.find(i))

ans=power(n,k,1000000000+7)
for i in s:
    p=graph.array[i][2]
    ans-=(power(p,k,1000000000+7))

print(ans%(1000000000+7))
