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. 

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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.
def dfs(edges,visited,node):
    red = 0
    if node not in visited:
        for kid in edges[node]:
            if kid[1] == 0 and kid[0] not in visited:
                red += 1
                break
            
    stack = [node]
    while stack:
        curr = stack.pop()
        if curr not in visited:
            found = False
            visited.add(curr)
            for kid in edges[curr]:
                if kid[1] == 0 and kid[0] not in visited:
                    stack.append(kid[0])
                    red += 1

    return red

def solve(edges,k):
    mod = 10**9+7
    n = len(edges)
    total = pow(n,k,mod)
    remove = 0
    visited = set()
    for i in range(1,n+1):
        if i not in visited:
            red = dfs(edges,visited,i)
            #print(red)
            remove += pow(red,k,mod)
            remove -= red
            remove = remove%mod

    remove += n
    remove = remove%mod
            
    print(total-remove)
            

def main():
    n,k = map(int,input().split())
    edges = {}
    for i in range(1,n+1):
        edges[i] = []

    for i in range(n-1):
        u,v,x = map(int,input().split())
        edges[u].append((v,x))
        edges[v].append((u,x))

    solve(edges,k)


main()
