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.
#pragma gcc optimize("Ofast")
#pragma loop_opt(on)
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define debug(x) 1&&(cout<<#x<<':'<<x<<'\n')
#define mkp make_pair
#define pb emplace_back
#define pf emplace_front
#define ff first
#define ss second
#define siz(v) (ll(v.size()))
#define all(v) begin(v),end(v)
#define REP(i,l,r) for(int i=(l);i<(r);i++)
#define mid (l+(r-l>>1))

using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> rank_tree;
constexpr long double PI = acos(-1);
constexpr ll INF = 1e18, MOD = 1000000007, N = 100000;

ll n,k,u,v,x,res=0;
ll par[N+1],sz[N+1];
void init(){iota(par,par+n,0),fill(sz,sz+n,1);}
ll anc(ll x){return x==par[x]?x:(par[x]=anc(par[x]));}
void join(ll x,ll y){
    x = anc(x), y = anc(y);
    if(x == y)return;
    if(sz[x] > sz[y])swap(x,y);
    sz[y] += sz[x], par[x] = y;
}
ll modmul(ll a,ll b){
    ll r=0;
    while(b){
        if(b&1)r+=a,r>=MOD&&(r-=MOD);
        a<<=1,a>=MOD&&(a-=MOD),b>>=1;
    }
    return r;
}
ll modpow(ll e,ll p){
    ll r=1;
    while(p){
        if(p&1)r=modmul(r,e);
        e=modmul(e,e),p>>=1;
    }
    return r;
}
signed main(){
    cin >> n >> k;
    init();
    REP(i,0,n-1){
        cin >> u >> v >> x, --u, --v;
        if(!x)join(u,v);
    }
    REP(i,0,n){
        u = anc(i);
        if(sz[u])res += modpow(sz[u],k), sz[u] = 0;
    }
    res = (MOD+modpow(n,k) - res)%MOD;
    cout << res << '\n';
}
