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.
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline


class UnionFind:
    def __init__(self, size):
        """
        size:
        parent[x]->parent of x
        rank[x] -> height of tree of x
        conN[x] ->size of group x belong
        """
        self.parent = [i for i in range(size)]
        self.rank = [0 for _ in range(size)]
        self.conN = [1 for _ in range(size)]

    def find(self, x):
        """
        """
        if self.parent[x] == x:
            return x
        else:
            return self.find(self.parent[x])

    def union(self, x, y):
        """
        """
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return

        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
            self.conN[y] += self.conN[x]
        else:
            self.parent[y] = x
            self.conN[x] += self.conN[y]
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

    def same(self, x, y):
        """
        """
        return self.find(x) == self.find(y)

    def component(self):
        """
        """
        comp = set()
        for i in self.parent:
            p = self.find(i)
            comp.add(p)
        return comp

    def componentNumm(self):
        """
        """
        return len(self.component())

    def connectedNum(self, x):
        """
        """
        p = self.find(x)
        return self.conN[p]

    def __str__(self):
        """
        for debug
        """
        ret = "parents\n"
        ret += " ".join(map(str, self.parent))
        ret += '\n'
        ret += " ".join(map(str, self.rank))
        return ret


MOD = 10**9+7
N, K = map(int, input().split())
Un = UnionFind(N)
for _ in range(N-1):
    a, b, c = map(int, input().split())
    if c == 0:
        a -= 1
        b -= 1
        Un.union(a, b)

noGood = 0
parent = set()
for i in range(N):
    p = Un.find(i)
    if p in parent:
        continue
    else:
        parent.add(p)
        noGood += pow(Un.connectedNum(p), K, MOD)
print(pow(N, K, MOD)-noGood)
