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 __future__ import division, print_function
py2 = round(0.5)

if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import BytesIO, IOBase

# FastIO in pypy =D /pajenegod

class FastI:
    def __init__(self, fd=0, buffersize=2**14):
        self.stream, self.bufflines = BytesIO(), 0
        def read2buffer(stream):
            curpos,s = stream.tell(), os.read(fd, buffersize + os.fstat(fd).st_size)
            return [s, stream.seek(0,2), stream.write(s), stream.seek(curpos)][0]
        self.read2buffer = lambda: read2buffer(self.stream)
        self.input = lambda: self.readline().rstrip(b'\r\n')
    
    def read(self):
        while self.read2buffer(): pass
        return self.stream.read()

    def readline(self):
        while self.bufflines == 0: 
            s = self.read2buffer()
            self.bufflines += s.count(b'\n') + (not s)
        self.bufflines -= 1
        return self.stream.readline()

class FastO(IOBase):
    def __init__(self, fd=1):
        stream = BytesIO()
        self.flush = lambda: os.write(fd, stream.getvalue()) + stream.truncate(0)
        self.write = stream.write if py2 else lambda s: stream.write(s.encode())

sys.stdin, sys.stdout = FastI(), FastO()
input = sys.stdin.input

################################# Testing filo code

def readLine():
    return [int(i) for i in input().split()]

def deg(x, y):
    if y == 1:
        return x
    res = deg(x, y // 2)
    res *= res
    res %= md
    if y % 2 == 1:
        res *= x
        res %= md
    return res

def dfs(v):
    Q = [v]
    used[v] = True
    c = 0
    while Q:
        v = Q.pop()
        c += 1
        for to in g[v]:
            if not used[to]:
                used[to] = True
                Q.append(to)
    return c

n, m = readLine()
g = [[] for _ in range(n + 1)]
md = 1000000007

for i in range(n - 1):
    x, y, t = readLine()
    if t == 0:
        g[x].append(y)
        g[y].append(x)

ans = deg(n, m)

used = [False]*(n+1)
for i in range(1, n + 1):
    if used[i] == False:
        b = dfs(i)
        ans = ans - deg(b, m)
        if ans < 0:
            ans += md
print(ans)