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.
# AC
import sys


class E:
    def __init__(self, fr, to, c):
        self.fr = fr - 1
        self.to = to - 1
        self.c = c


class Main:
    def __init__(self):
        self.buff = None
        self.index = 0

    def next(self):
        if self.buff is None or self.index == len(self.buff):
            self.buff = sys.stdin.readline().split()
            self.index = 0
        val = self.buff[self.index]
        self.index += 1
        return val

    def next_int(self):
        return int(self.next())

    def solve(self):
        n = self.next_int()
        k = self.next_int()
        fa = [-1 for _ in range(0, n)]
        sz = [1 for _ in range(0, n)]
        for _ in range(1, n):
            fr = self.find(self.next_int() - 1, fa)
            to = self.find(self.next_int() - 1, fa)
            c = self.next_int()
            if c == 0 and fr != to:
                sz[fr] += sz[to]
                fa[to] = fr
        mod = 1000000007
        ans = self.pp(n, k, mod)
        for i in range(0, n):
            if fa[i] == -1:
                ans = (ans + mod - self.pp(sz[i], k, mod)) % mod
        print(ans)

    def find(self, k, fa):
        if fa[k] == -1:
            return k
        fa[k] = self.find(fa[k], fa)
        return fa[k]

    def pp(self, n, k, mod):
        if k == 0:
            return 1
        tmp = self.pp(n, k // 2, mod)
        tmp = tmp * tmp % mod
        if k % 2 == 1:
            tmp = tmp * n % mod
        return tmp


if __name__ == '__main__':
    Main().solve()
