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.
import sys
input=sys.stdin.readline
class disjoint:
    def __init__(self,n):
        self.rank=[1]*n
        self.parent=[i for i in range(n)]
       
        
    def find(self,x):
        if(self.parent[x]!=x):
            self.parent[x]=self.find(self.parent[x])
        return self.parent[x];
       
    def union(self,x,y):
        xid=self.find(x)
        yid=self.find(y)
        
        if(xid==yid):
            return;
        if(self.rank[xid]<self.rank[yid]):
            self.parent[xid]=yid
            
            return;
        elif(self.rank[xid]>self.rank[yid]):
            self.parent[yid]=xid # merging y into x
          
            return;
        else:
            self.parent[yid]=xid
            self.rank[xid]+=1
            
            return;        
    
from collections import *
obj=disjoint((10**5)+1)
pri=pow(10,9)+7
n,k=map(int,input().split())
for i in range(n-1):
    x,y,z=map(int,input().split())
    if(z==0):
        obj.union(x,y);

    
al=defaultdict(int)
total=0
for i in range(1,n+1):
    if(al[obj.find(i)]==0):
        total+=pow(obj.rank[i],k,pri)
        al[obj.find(i)]
print((pow(n,k,pri)-total)%pri)


#init function sort of creates the parent and the rank list
#find function is used for to find rank
#union function is used for doing the union
#what does path function do , it makes find operation almost linear
#makes our tree of very less height


#use of log *n function -

#number times log has to be applied to make log value as 1

#by using path compression time complexty of each function is log*n

#
    
            
                            
    
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