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
from collections import *


class graph:
    # initialize graph
    def __init__(self, gdict=None):
        self.gdict = [[] for _ in range(10 ** 5 + 1)]

    # add edge
    def addEdge(self, node1, node2, w=None):
        self.gdict[node1].append(node2)
        self.gdict[node2].append(node1)

    def bfs_util(self, i):
        queue, self.visit[i], nodes = deque([i]), 1, 0

        while queue:
            # dequeue parent vertix
            s = queue.popleft()
            nodes += 1

            # enqueue child vertices
            for v in self.gdict[s]:
                if self.visit[v] == 0:
                    self.visit[v] = 1
                    queue.append(v)

        return nodes

    def bfs(self):
        self.visit, self.cnt = [0] * (n + 1), 0

        for i in range(1, n + 1):
            if self.visit[i] == 0:
                self.cnt = add(self.cnt, pow(self.bfs_util(i), k, mod))

        print(add(pow(n, k, mod), -self.cnt))


add = lambda a, b: (a % mod + b % mod) % mod
mult = lambda a, b: ((a % mod) * (b % mod)) % mod
rints = lambda: [int(x) for x in stdin.readline().split()]
g, mod = graph(), 10 ** 9 + 7

n, k = rints()
for _ in range(n - 1):
    u, v, w = rints()
    if w == 0:
        g.addEdge(u, v)

g.bfs()
