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.
from collections import defaultdict as dd
mod= pow(10,9)+7
"""
def dfs_visit(V,adj,s,c):
    for v in adj[s]:
        if v not in parent:
            parent[v]=s
            c=c+1
            dfs_visit(V,adj,v,c)
    return(c)
            
def dfs(V,adj):
    ans=[]
    for s in V:
        if s not in parent:
            parent[s]=None
            c=dfs_visit(V,adj,s,1)
            ans.append(c)
    return(ans)
"""
def bfs(s,adj):
    global parent
    t=1
    #print("df",s)
    frontier=[s]
    while frontier:
        nex=[]
        for u in frontier:
            for v in adj[u]:
                if v not in parent:
                    #print(v)
                    t+=1
                    parent.add(v)
                    nex.append(v)
        frontier=nex
    return(t)

adj=dd(list)
n,k=[int(i) for i in input().split(' ')]
for ii in range(n-1):
    x,y,h=[int(i) for i in input().split(' ')]
    if h==0:
        adj[x].append(y)
        adj[y].append(x)

V=[i+1 for i in range(n)]  
parent=set()

fk=[]
for i in V:
    if i not in parent:
        parent.add(i)
        tr=bfs(i,adj)
        #print(i,tr)
        fk.append(tr)
#print(fk)
ans=pow(n,k,mod)
for i in fk:
    ans=(ans-pow(i,k,mod))%mod
print(ans%mod)
