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
M = 10**9 + 7
class Graph:
    def __init__(self,n,k):
        self.red =defaultdict(list)
        self.black = defaultdict(list)
        self.visitedRed = defaultdict(bool)
        self.vertices = n
        self.power = k
        self.minus = 0
        self.blackPowerNegative = 0
    def addEdge(self,fr,to,col):
        if(col==0):
            self.red[fr].append(to)
            self.red[to].append(fr)
        else:
            self.black[fr].append(to)
            self.black[to].append(fr)
    def BFSUtil(self,root):
        count = 0
        if(len(self.red[root])==0):
            return
        queue = [root]
        self.visitedRed[root] = True
        while(queue):
            s = queue.pop(0)
            count += 1
            for k in self.red[s]:
                if(self.visitedRed[k]==False):
                    self.visitedRed[k] = True
                    queue.append(k)
        self.minus+=pow(count,self.power,M)
    def BFS(self):
        for i in range(1,self.vertices+1):
            if(self.visitedRed[i]==False):
                self.BFSUtil(i)
    def BFSBlack(self):
        for i in self.black.keys():
            if(not self.red[i]):
                self.blackPowerNegative+=1
    def solve(self):
        # print("S: ",self.minus)
        X = (pow(self.vertices,self.power,M) - self.minus)%M - self.blackPowerNegative
        print(X)
n,k = map(int,input().split())
G = Graph(n,k)
for i in range(n-1):
    a,b,col = [int(x) for x in input().split()]
    G.addEdge(a,b,col)
G.BFS()
G.BFSBlack()
G.solve()