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 math,sys
#from itertools import permutations, combinations;import heapq,random;
# from collections import defaultdict,deque
import bisect as bi
def yes():print('YES')
def no():print('NO')
#sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(sys.stdin.readline()))
def In():return(map(int,sys.stdin.readline().split()))
def Sn():return sys.stdin.readline().strip()
#sys.setrecursionlimit(1500)
def dict(a):
    d={} 
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d
def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_left(a, x)
    if i != len(a):
        return i
    else:            
        return -1
def dfs(d,visit,s):
    q=[s]
    cnt=0
    while q:
        t=q.pop(0)
        visit[t]=True
        cnt+=1
        for x in d[t]:
            if visit[x]==False:
                q.insert(0,x)
    return cnt
def mod(a, b, c):
    res = a * b
    return res % c

def power(a, b, m):
    x, y = 1,a
    while (b > 0):
        if (b & 1):x = mod(x, y, m)
        y = mod(y, y, m)
        b >>= 1
    return x
def expo(a, b):
    x, y = 1, a
    while (b > 0):
        if (b & 1):
            x = x * y
        y = y * y
        b >>= 1
    return x
def main():
    try:
        n,k=In()
        d=[[] for x in range(n+1)]
        for i in range(n-1):
            a,b,c=In()
            if c==0:
                d[a].append(b)
                d[b].append(a)
        ans=power(n,k,P)
        visit=[False for x in range(n+1)]
        for i in range(1,n+1):
            if not visit[i]:
                ans-=power(dfs(d,visit,i),k,P)
        print(ans%P)
    except:
        pass

M = 998244353
P = 1000000007
 
if __name__ == '__main__':
    # for _ in range(I()):main()
    for _ in range(1):main()
#End#

#        ******************* All The Best *******************   #