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.
import java.util.*;

public final class c{
	static int n,k;
	static LinkedList<Integer> adj[];
	static LinkedList<Integer> components = new LinkedList();
	static boolean visited[];
	static int mod = 1000000007;

	public static int DFS(int v){
		int cnt = 0;
		visited[v] = true;
		for(int u : adj[v]){
			if(!visited[u]){
				cnt += DFS(u);
			}
		}
		return cnt+1;
	}

	public static long power(long n, long k){
		// System.out.println(n + " " + k);
		if(k == 0){
			return 1;
		}
		long x = power(n,k/2);
		if(k%2 == 0){
			return (x*x)%mod;
		}
		else{
			return (((x*x)%mod)*n)%mod;
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		k = sc.nextInt();
		adj = new LinkedList[n];
		visited = new boolean[n];
		for(int i = 0;i < n;i++){
			adj[i] = new LinkedList<Integer>();
		}
		for(int i = 0;i < n-1;i++){
			int u = sc.nextInt();
			int v = sc.nextInt();
			int x = sc.nextInt();
			if(x == 0){
				adj[u-1].add(v-1);
				adj[v-1].add(u-1);
			}
		}

		for(int i = 0;i < n;i++){
			if(!visited[i]){
				components.add(DFS(i));
			}
		}


		long result = 0;
		
		for(int j : components){
			result = (result + power((long)j,(long)k))%mod;
		}
		System.out.println(power((long)n,(long)k) + " " +  result);
		result = ((power((long)n,(long)k) - result)%mod)%mod;
		System.out.println(result);
	}
}