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.ArrayList;
import java.util.Scanner;
import java.util.*;
import static java.lang.Math.*;


public class java2 {
    static void dfs(int u)
    {
        check[u]=1;
        ++cnt;
        for(int x:vec[u])
        {
            if(check[x]==0) dfs(x);
        }
    }
    static long fastPow(long a,long b)
    {
        if(b==0) return 1L;
        long res=fastPow(a,b/2);
        if(b%2==0) return res*res%mod;
        else return res*res%mod*a%mod;
    }
    static long mod=(long)1e9+7;
    static ArrayList<Integer> vec[];
    static int check[];
    static long cnt=0;
    public static void main(String[] args) {
        Scanner r = new Scanner(System.in);
        int n=r.nextInt();
        int k=r.nextInt();
        vec=new ArrayList[n];
        for(int i=0;i<n;++i) vec[i]=new ArrayList<>();
        for(int i=0;i<n-1;++i)
        {
            int u=r.nextInt()-1;
            int v=r.nextInt()-1;
            int x=r.nextInt();
            if(x==0)
            {
                vec[u].add(v);
                vec[v].add(u);
            }
        }
        long ans=fastPow(n,k);
        long res=0;
        check = new int[n];
        for(int i=0;i<n;++i)
        {
            if(check[i]==0)
            {
                cnt=0;
                dfs(i);
                res+=fastPow(cnt,k);
            }
        }
        res%=mod;
        ans=(ans-res+mod)%mod;
        System.out.println(ans);
    }
}
