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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class Main {
	static int[] par;
	static int[] len;
	static long mod = 1000000007;

	public static void main(String[] args) {
		Reader in = new Reader();
		int n = in.nextInt();
		int k = in.nextInt();
		par = new int[n+1];
		len = new int[n+1];
		for(int i = 0; i<=n; i++) {
			par[i] = i;
			len[i] = 1;
		}
		long ans = power(n, k, mod);
		for(int i = 0; i<n-1; i++) {
			int a = in.nextInt();
			int b = in.nextInt();
			int c = in.nextInt();
			if(c==0) unite(a, b);
		}
		for(int i = 1; i<=n; i++) {
			if(par[i]==i) {
				ans-=power(len[i], k, mod);
			}
		}
		System.out.println(ans);
	}

	public static int find(int x) {
		if (par[x] == x)
			return x;
		return find(par[x]);
	}

	public static void unite(int a, int b) {
		int p1 = find(a);
		int p2 = find(b);
		par[p2] = p1;
		len[p1] += len[p2];
	}

	static long power(long x, long y, long p) {
		long res = 1;
		x = x % p;
		while (y > 0) {
			if ((y & 1) == 1)
				res = (res * x) % p;
			y = y >> 1;
			x = (x * x) % p;
		}
		return res;
	}

	static class Reader {
		static BufferedReader br;
		static StringTokenizer st;

		public Reader() {
			this.br = new BufferedReader(new InputStreamReader(System.in));
		}

		public int[] na(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public long[] nl(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public String[] nS(int n) {
			String[] a = new String[n];
			for (int i = 0; i < n; i++)
				a[i] = next();
			return a;
		}

		public int nextInt() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return Integer.parseInt(st.nextToken());
		}

		public double nextDouble() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return Double.parseDouble(st.nextToken());
		}

		public Long nextLong() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return Long.parseLong(st.nextToken());
		}

		public String next() {
			if (st == null || !st.hasMoreTokens())
				try {
					readLine();
				} catch (Exception e) {
				}
			return st.nextToken();
		}

		public static void readLine() {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
			}
		}
	}
}