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.
#include<bits/stdc++.h>
#include <ext/numeric> 
using namespace std;
using namespace __gnu_cxx;
#define D(x) cout << #x " is " << x << endl
#define fi first
#define se second
typedef long long lng;
typedef pair<lng, lng> pii;
typedef pair<lng, pii> piii;

const lng N = 1e5+5;
lng head[N], to[N], nxt[N];
bool vis[N];
lng n, k, ne;

const lng MOD = 1e9 + 7;
struct MUL{
	lng operator ()(lng a, lng b) const {
		return (1LL * a * b) % MOD;		
	}
};

lng identity_element(const MUL &) {
	return 1;
}

const MUL mul;

lng dfs(lng u){
	if(vis[u])
		return 0;
	vis[u] = true;
	lng res = 1;
	for(lng k = head[u]; ~k; k = nxt[k]){
		lng v = to[k];
		res += dfs(v);
	}
	return res;
}

void init(){
	ne = 0;
	memset(head, -1, n*sizeof head[0]);
}

void add_edge(lng f, lng t){
	to[ne] = t;
	nxt[ne] = head[f];
	head[f] = ne++;
}
void add_biedge(lng f, lng t){
	add_edge(f, t);
	add_edge(t, f);
}

int main(){
	scanf("%lld%lld", &n, &k);
	init();
	lng u, v, c;
	for(lng i = 0; i < n-1; ++i){
		scanf("%lld%lld%lld", &u, &v, &c);
		--u, --v;
		if(!c)
			add_biedge(u, v);
	}
	
	lng ans = 0;
	for(lng i = 0; i < n; ++i){
		lng m = 0;
		if(!vis[i])
			m = dfs(i);
		ans += power(m, k, mul);
		ans %= MOD;
	}
	
	lng tot = power(n, k, mul);
	ans = tot - ans;
	
	printf("%lld", ans);
	
	return 0;
}
