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.List;
import java.util.Scanner;

public class _1139C {

    public static final long MOD = 1000000007l;
    Scanner scanner = new Scanner(System.in);
    int n, k;
    UnionFind unionFind;

    public static void main(String[] args) {
        _1139C solution = new _1139C();

        solution.solve();
    }

    private class UnionFind{
        int[] parent = new int[n+1];
        int[] weight = new int[n+1];

        {
            for(int i=1; i<=n; i++){
                parent[i] = i;
                weight[i] = 1;
            }
        }

        private int find(int v){
            if(parent[v]!=v){
                parent[v] = find(parent[v]);
                return parent[v];
            }
            return v;
        }

        private void union(int u, int v){
            int pu = find(u), pv = find(v);

            if(pu==pv) return;

            if(weight[pu] > weight[pv]){
                parent[pv] = pu;
                weight[pu] += weight[pv];
            }
            else{
                parent[pu] = pv;
                weight[pv] += weight[pu];
            }
        }
    }

    public void solve(){
        n = scanner.nextInt();
        k = scanner.nextInt();
        unionFind = new UnionFind();


        int u, v, col;
        for(int i=1; i<=n-1; i++){
            u = scanner.nextInt();
            v = scanner.nextInt();
            col = scanner.nextInt();

            if(col==0) {
                unionFind.union(u, v);
            }
        }

        System.out.println(pathWithBlack());
    }

    private long pathWithBlack(){
        List<Long> groupSize = new ArrayList<>();
        long noOfPaths = powK(n, k);

        for(int i=1; i<=n; i++){
            if(unionFind.parent[i] == i){
                groupSize.add((long) unionFind.weight[i]);
            }
        }

        for(long grp: groupSize){
            noOfPaths = subtractWithModulo(noOfPaths, powK(grp, k));
        }

        return noOfPaths;
    }

    private long subtractWithModulo(long a, long b){
        return ((a-b) % MOD + MOD) % MOD;
    }

    private long powK(long n, long k){
        long ans=1;
        long multiplier = n;
        while(k>0){
            if(k%2==1){
                ans = (ans * multiplier) % MOD;
            }
            multiplier = (multiplier * multiplier) % MOD;
            k /= 2;
        }

        return ans;
    }


}