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.
#import resource
import sys
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
import threading
threading.stack_size(2**26)
sys.setrecursionlimit(10**7)
mod=(10**9)+7
#fact=[1]
#for i in range(1,1001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*1001
#ifact[1000]=pow(fact[1000],mod-2,mod)
#for i in range(1000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
from sys import stdin, stdout
#from bisect import bisect_left as bl
#from bisect import bisect_right as br
#import itertools
#import math
#import heapq
#from random import randint as rn
#from Queue import Queue as Q
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def ain():
    return map(int,sin().split())
def sin():
    return stdin.readline().strip()
def GCD(x, y):
   while(y):
       x, y = y, x % y
   return x
"""**************************************************************************"""
def main():
    def dfs(x):
        if(p[x]==0):
            p[x]=1
            k[0]+=1
            for i in adj[x]:
                dfs(i)
    n,k1=ain()
    adj=[[] for i in range(n)]
    for i in range(n-1):
        x,y,w=ain()
        if(w==0):
            adj[y-1].append(x-1)
            adj[x-1].append(y-1)
    p=[0]*n
    k=[0]
    t=[]
    for i in range(n):
        if(p[i]==0):
            k[0]=0
            dfs(i)
            t.append(k[0])
    s=pow(n,k1,mod)
    f=0
    for i in range(len(t)):
        f=(f+pow(t[i],k1,mod))%mod
    s=(s-f)%mod
    print s
threading.Thread(target=main).start()
