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.
#include <iostream>
#include <stack>
#include <cmath>

using namespace std;

short graph[10001][10001];
long long otv = 0;
long long otvmax = 0;
long long search = 0;
int n;

void dfs(int tmp1,int tmp2, stack <int> &a) {
	for (int i = 0; i <= n; i++) {
		if (i != tmp2 && i!= tmp1) {
			if (graph[tmp1][i] == 1) {
				search++;
				dfs(i, tmp1, a);
			}
			if (graph[tmp1][i] == 2) {
				a.push(tmp1); a.push(i);
			}
		}
	}
}

void check(long long &t) {
	if (t >= 1000000007) {
		t %= 1000000007;
	}
}

int main() {
	stack <int> a;
	int  k;
	cin >> n >> k;
	otvmax = 1;
	for (int i = 0; i < k; i++) {
		otvmax *= n;
		check(otvmax);
	}

	int tmp1, tmp2, tmp3;
	a.push(-1);
	a.push(-1);
	bool first = true;
	for (int i = 0; i < n-1; i++) {
		cin >> tmp1 >> tmp2 >> tmp3;
		graph[tmp1][tmp2] = tmp3 + 1;
		graph[tmp2][tmp1] = tmp3 + 1;
		graph[tmp2][tmp2] = 1;
		graph[tmp1][tmp1] = 1;
		if (tmp3 == 1 && first) {
			a.push(tmp1); a.push(tmp2);
			a.push(tmp2); a.push(tmp1);
			first = false;
		}
	}
	if (a.size() == 2) {
		cout << 0;
		//system("pause");
		return 0;
	}
	long long left, right;
	while (true) {
		tmp1 = a.top(); a.pop();
		tmp2 = a.top(); a.pop();
		if (tmp1 == -1) break;
		search = 1;
		dfs(tmp1, tmp2, a);
		check(otv);
		long long temp = 1;
		for (int i = 0; i < k; i++) {
			temp *= search;
			check(temp);
		}
		otv += temp;
		check(otv);
	}
	check(otv);
	cout << otvmax - otv;
	//system("pause");
	return 0;
}