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 sys import stdin,stdout

n,k=stdin.readline().strip().split(' ');n,k=int(n),int(k)
adj=[[] for i in range(n)]
ver_col=[[] for i in range(n)]
for i in range(n-1):
	u,v,x=stdin.readline().strip().split(' ');u,v,x=int(u),int(v),int(x)
	adj[u-1].append(v-1);ver_col[u-1].append(x)
	adj[v-1].append(u-1);ver_col[v-1].append(x)



count=[]
# nckarr=[[0,0] for i in range(n-k+2)]
# def fillnckarr():
# 	global nckarr,n,k
# 	nckarr[0][0]=k;
# 	nckarr[0][1]=1;

# 	for i in range(1,len(nckarr)):
# 		nckarr[i][1]=nckarr[i-1][0]+nckarr[i-1][1];
# 		nckarr[i][0]=int(nckarr[i][1]*k/(i+1))

# fillnckarr()

def getnck(n):
	global nckarr
	if n<k:
		return 0
	else:
		return nckarr[n-k][1]


def dfs(current_vertex,parent_vertex,idd):
	global count
	if len(count)==idd:
		count.append(0)
		count[idd]+=1
	else:
		count[idd]+=1

	for i in range(len(adj[current_vertex])):
		if adj[current_vertex][i]!=parent_vertex:
			if ver_col[current_vertex][i] == 1:
				dfs(adj[current_vertex][i],current_vertex,idd+1)
			else:
				dfs(adj[current_vertex][i],current_vertex,idd)

dfs(0,-1,0)

#print(count)
						
# for i in nckarr:
# 	print(i)
mo=10**9 + 7
ans=pow(n,k,mo)

for i in count:
	ans=(ans-pow(i,k,mo))%mo
stdout.write(str(ans)+"\n")




