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.*;
import java.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
    
    static int n,k,ns;
    static ArrayList<Integer>[]A;
    static int[]S;
    static long ans;
    static long m=1000000007;
    
    void solve() throws IOException {
        n=ni();
        k=ni();
        A=new ArrayList[n+1];
        S=new int[n+1];
        for (int x=1;x<=n;x++) A[x]=new ArrayList();
        for (int x=1;x<n;x++) {
            int u=ni(),v=ni(),w=ni();
            A[u].add(v);
            A[u].add(w);
            A[v].add(u);
            A[v].add(w);
        }
        
        ans=mp(n,k);
        ns=1;
        dfs(1,0,1);
        
        for (int x=1;x<=ns;x++) {
            ans=(ans-mp(S[x],k))%m;
            if (ans<0) ans=ans+m;
        }
        
        System.out.println(ans);
    }
    
    long mp(long a,long b) {
        long r=1;
        while (b>0) {
            if ((b&1)>0) r=(r*a)%m;
            b=b>>1;
            a=(a*a)%m;
        }
        return r;
    }
    
    void dfs(int v,int p,int s) {
        S[s]++;
        for (int x=0;x<A[v].size();x+=2) {
            int u=A[v].get(x);
            int w=A[v].get(x+1);
            if (u==p) continue;
            if (w==0) dfs(u,v,s);
            else { ns++; dfs(u,v,ns); }
        }
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}