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.
# curr_comp
# dfs(v):
#     for all children of v
#         if not visited c
#             visited c = true
#             comp c = curr comp
#             dfs(c)
#
# dfs(v):
#     if visited v:
#         return
#     visited v = true
#     comp v = curr
#
#     for all red children:
#         dfs(c)
import functools

if __name__ == '__main__':
    nodes_nr, seq_len = (int(x) for x in input().split())
    node_idx___red_children = [[] for _ in range(nodes_nr)]
    mod = 10 **9 + 7
    for _ in range(nodes_nr - 1):
        n1, n2, color = (int(x) for x in input().split())
        if color == 0:
            node_idx___red_children[n1 - 1].append(n2 - 1)
            node_idx___red_children[n2 - 1].append(n1 - 1)

    node_idx___component = [-1 for _ in range(nodes_nr)]
    component___size = []
    curr_component = 0
    for node in range(nodes_nr):
        if node_idx___component[node] != -1:
            continue

        curr_component_size = 1

        stack = [node]
        node_idx___component[node] = curr_component
        while len(stack) > 0:
            node_idx = stack.pop()
            for child_idx in node_idx___red_children[node_idx]:
                if node_idx___component[child_idx] == -1:
                    node_idx___component[child_idx] = curr_component
                    curr_component_size += 1
                    stack.append(child_idx)
                elif node_idx___component[child_idx] != curr_component:
                    raise Exception('Two connectivity components are connected somehow. Go investigate it')

        component___size.append(curr_component_size)
        curr_component += 1

    ans = pow(nodes_nr, seq_len, mod)
    for size in component___size:
        ans -= pow(size, seq_len, mod)
        ans %= mod
    print((ans % mod + mod) % mod)