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.*;
import static java.lang.System.in;

public class Main {
    static int n,k;
    static long mod = 1000000000+7;
    static long[] fac,inv;
    static int[] id,size;
    //static long mod = 1000000000+7;
    //static HashSet<Integer>[] graph;
    public static void main(String[] args)throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(in));
        String[] buf = reader.readLine().split(" ");
        //Scanner sc = new Scanner(System.in);
        n = Integer.parseInt(buf[0]); k = Integer.parseInt(buf[1]);
        init();
        id = new int[n+1]; size = new int[n+1];
        Arrays.fill(size,1);
        for(int i=1;i<=n;i++) id[i] = i;
        for(int i=0;i<n-1;i++){
            buf = reader.readLine().split(" ");
            int a=Integer.parseInt(buf[0]), b=Integer.parseInt(buf[1]), color=Integer.parseInt(buf[2]);
            if(color==0) unite(a,b);
        }
        boolean[] vis = new boolean[n+1];
        long res = 0; // to record the only-red-edge choices
        for(int i=1;i<=n;i++){
            int root = find(i);
            if(vis[root]) continue;
            vis[root] = true;
            int len = size[root];
            long cur = power(len,k)-len;
            res = (res+cur)%mod;
        }
        long ans = power(n,k)-n;
        ans = (ans+mod-res)%mod;
        System.out.println(ans);
    }
    static int find(int p){
        while(p!=id[p]){
            id[p]=id[id[p]];
            p=id[p];
        }
        return p;
    }
    static void unite(int p, int q){
        int i=find(p), j=find(q);
        if(i==j) return;
        id[j] = i;
        size[i] += size[j];
    }
    static void init(){
        fac = new long[n+1]; inv = new long[n+1];
        fac[0] = 1; inv[0] = 1;
        for(int i=1;i<=n;i++) fac[i] = fac[i-1]*i;
        inv[n] = power(fac[n],mod-2);
        for(int i=n-1;i>=0;i--) inv[i] = (i+1)*inv[i+1]%mod;
    }
    static long power(long base, long p){
        long ans = 1;
        while(p>0){
            if(p%2==1) ans = ans*base%mod;
            base = base*base%mod;
            p /= 2;
        }
        return ans;
    }
}