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.io.*;
public class Main{
    static int faultNodes=0;
    public static void main(String args[])throws IOException{
        BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
        String inp[]=br.readLine().trim().split(" ");
        int n=Integer.parseInt(inp[0]);
        int k=Integer.parseInt(inp[1]);
        ArrayList<Integer> arr[]=new ArrayList[n+1];
        for(int i=0;i<=n;i++) arr[i]=new ArrayList<>();
        boolean fault[]=new boolean[n+1];
        int mod=1000000007;
        for(int i=0;i<n+1;i++) fault[i]=true;
        for(int i=0;i<n-1;i++){
            inp=br.readLine().trim().split(" ");
            int a=Integer.parseInt(inp[0]);
            int b=Integer.parseInt(inp[1]);
            int col=Integer.parseInt(inp[2]);
            arr[a].add(b);
            arr[b].add(a);
            if(col == 1) fault[a]=fault[b]=false;
        }
        boolean visited[]=new boolean[n+1];
        int totalFaultyNodes=0;
        long toSub=0;
        for(int i=1;i<=n;i++){
            if(!visited[i] && fault[i]){
                faultNodes=0;
                dfs(i,arr,visited,fault);
                totalFaultyNodes+=faultNodes+1;
                long temp=binexp(faultNodes+1,k,mod);
                toSub=(toSub%mod + temp%mod)%mod;
            }
        }
        toSub=(toSub%mod + (n-totalFaultyNodes)%mod)%mod;
        
        long ans=binexp(n,k,mod)%mod-toSub%mod;
        if(ans<0) ans=0;
        System.out.println(ans);
    }
    
    static void dfs(int i,ArrayList<Integer> arr[],boolean visited[],boolean fault[]){
        visited[i]=true;
        faultNodes++;
        int len=arr[i].size();
        for(int j=0;j<len;j++){
            int index=arr[i].get(j);
            if(!visited[index] && fault[index])
            dfs(index,arr,visited,fault);
        }
    }
    static long binexp(long  a,  long b,  long m) {
    a %= m;
     long res = 1;
    while (b > 0) {
        if (b%2 == 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}
}