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. 

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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>
using namespace std;
#define ll long long

const ll N = 100001;
std::vector<ll> g[N];
ll M = (long long)10e9+7;
ll used[N] = {0};

ll dfs(ll n,ll count_comp)
{
	if(!g[n].empty())
		used[n] = 1;
	
	for(ll v:g[n]) {
		if(!used[v]) {
			count_comp++;
			count_comp = dfs(v,count_comp);
		}
	}
	return count_comp;
}

ll solve()
{
	ll n,k;
	cin>>n>>k;
	for(ll i =1;i<n;i++) {
		ll x,y,col;
		cin>>x>>y>>col;
		if(col == 0) {
			g[x].push_back(y);
			g[y].push_back(x);
		}
	}
	ll connect[n+1] = {0};
	ll count_comp;
	for(ll i=1;i<=n;i++) {
		if(!used[i]) {
			count_comp = 0;
			connect[i] = dfs(i,count_comp);
			// cout<<connect[i]<<endl;
		}
	}

	ll total = 0;

	for(ll i=1;i<=n;i++) {
		if(connect[i])
			total += pow(connect[i]+1,k);
	}
	for(ll i=1;i<=n;i++) {
		if(used[i] == 0) {
			total += 1;
		}
	}
	ll res = ((long long)pow(n,k) - total) % M;
	return res;

}

int main()
{
	int t = 1;
	// cin>>t;
	while(t--) 
		cout<<solve()<<endl;

}