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.
mod = 10**9+7
import sys
input = sys.stdin.buffer.readline
from collections import deque

class Graph(object):
	"""docstring for Graph"""
	def __init__(self,n,d): # Number of nodes and d is True if directed
		self.n = n
		self.graph = [[] for i in range(n)]
		self.parent = [-1 for i in range(n)]
		self.directed = d
		
	def addEdge(self,x,y):
		self.graph[x].append(y)
		if not self.directed:
			self.graph[y].append(x)

	def bfs(self, root, vis): # NORMAL BFS
		queue = [root]
		queue = deque(queue)
		c = 0
		while len(queue)!=0:
			element = queue.popleft()
			c += 1
			vis[element] = 1
			for i in self.graph[element]:
				if vis[i]==0:
					queue.append(i)
					self.parent[i] = element
		return c

	def dfs(self, root, ans): # Iterative DFS
		stack=[root]
		vis=[0]*self.n
		stack2=[]
		while len(stack)!=0: # INITIAL TRAVERSAL
			element = stack.pop()
			if vis[element]:
				continue
			vis[element] = 1
			stack2.append(element)
			for i in self.graph[element]:
				if vis[i]==0:
					self.parent[i] = element
					stack.append(i)

		while len(stack2)!=0: # BACKTRACING. Modify the loop according to the question
			element = stack2.pop()
			m = 0
			for i in self.graph[element]:
				if i!=self.parent[element]:
					m += ans[i]
			ans[element] = m
		return ans

	def shortestpath(self, source, dest): # Calculate Shortest Path between two nodes
		self.bfs(source)
		path = [dest]
		while self.parent[path[-1]]!=-1:
			path.append(parent[path[-1]])
		return path[::-1]

	def ifcycle(self):
		self.bfs(0)
		queue = [0]
		vis = [0]*n
		queue = deque(queue)
		while len(queue)!=0:
			element = queue.popleft()
			vis[element] = 1
			for i in graph[element]:
				if vis[i]==1 and i!=parent[element]:
					return True
				if vis[i]==0:
					queue.append(i)
					vis[i] = 1
		return False

	def reroot(self, root, ans):
		stack = [root]
		vis = [0]*n
		while len(stack)!=0:
			e = stack[-1]
			if vis[e]:
				stack.pop()
				# Reverse_The_Change()
				continue
			vis[e] = 1
			for i in graph[e]:
				if not vis[e]:
					stack.append(i)
			if self.parent[e]==-1:
				continue
			# Change_The_Answers()



n,k = map(int,input().split())
g = Graph(n,False)
for i in range(n-1):
	a,b,c = map(int,input().split())
	if not c:
		g.addEdge(a-1,b-1)
vis = [0]*n
count = 0
for i in range(n):
	if not vis[i]:
		c = g.bfs(i,vis)
		count += pow(c,k,mod)
		count %= mod
print ((pow(n,k,mod)-count)%mod)