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<bits/stdc++.h>
#define int long long
using namespace std;
int n,a,b,x,k,Ans,Mod=1e9+7,fix[1000001];
vector<int>v1[100001],v2[100001],V;
int dfs(int x)
{
	int ans = 0;
	if(fix[x]) return 0;
	fix[x] = 1;
	for(int i=0; i<v1[x].size(); i++)
		ans += dfs(v1[x][i]);
	return ++ans;
}
int fast_pow(int x, int y)
{
	if(y == 1) return x;
	int p = fast_pow(x,y/2) % Mod;
	if(y % 2)
		p = (p * p) % Mod * x % Mod;
	else
		p = (p * p) % Mod;
	return p;
}
main()
{
	cin >> n >> k;
	for(int i=1; i<n; i++)
	{
		cin >> a >> b >> x;
		if(x == 0)
		{
			v1[a].push_back(b);
			v1[b].push_back(b);
		}
	}
	for(int i=1; i<=n; i++)
	{
		if(!fix[i]) 
		{
			V.push_back(dfs(i));
		}
	}
	Ans = fast_pow(n,k);
	for(int i=0; i<V.size(); i++)
	{
		Ans -= fast_pow(V[i],k);
	}
	cout << (Ans + Mod) % Mod;
}
