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.
# -*- coding: utf-8 -*-
# @Time    : 2019/3/21 23:56
# @Author  : LunaFire
# @Email   : gilgemesh2012@gmail.com
# @File    : C. Edgy Trees.py


class UnionFind(object):
    def __init__(self, n):
        self.root = [i for i in range(n)]
        self.size = [1 for _ in range(n)]

    def find(self, x):
        if self.root[x] == x:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]

    def union(self, x, y):
        x, y = self.find(x), self.find(y)
        if x == y:
            return
        if self.size[x] < self.size[y]:
            x, y = y, x
        self.root[y] = x
        self.size[x] += self.size[y]


def pow_mod(a, b, p):
    ret = 1
    while b > 0:
        if b % 2 == 1:
            ret = ret * a % p
        b //= 2
        a = a * a % p
    return ret


def main():
    n, k = map(int, input().split())

    union_find = UnionFind(n + 1)
    for _ in range(n - 1):
        u, v, c = map(int, input().split())
        if c == 0:
            union_find.union(u, v)

    p = int(1e9 + 7)
    ret = pow_mod(n, k, p)
    used = set()
    for i in range(1, n + 1):
        x = union_find.find(i)
        if x not in used:
            ret = (ret + p - pow_mod(union_find.size[x], k, p)) % p
            used.add(x)
    print(ret)


if __name__ == '__main__':
    main()
    # print(pow_mod(4, 4, 3))