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>
#define ios ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long
#define bit(x,j) ((x>>j)&1)
using namespace std;
// bitset<8>set1(p);
// __builtin_popcountll(x);
// bitset<100> b("1010");
ll int mod=1e9+7;//998244353;
int mul(ll int x, ll int y)
{
    return (x * 1ll * y) % mod;
}

int add(int x, int y)
{
    x += y;
    while(x >= mod) x -= mod;
    while(x < 0) x += mod;
    return x;
}

long long power(long long a, long long b,ll m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = mul(res,a);
        a = mul(a,a);
        b >>= 1;
    }
    return res%m;
    }


  ll int binomial(ll int n, ll int k,ll m) 
{ 
    ll int C[k+1]; 
    memset(C, 0, sizeof(C)); 
  
    C[0] = 1;  // nC0 is 1 
  
    for (ll int i = 1; i <= n; i++) 
    { 
        // Compute next row of pascal triangle using 
        // the previous row 
        for (int j = min(i, k); j > 0; j--) 
            C[j] = add(C[j],C[j-1]); 
    } 
    return C[k]%m; 
} 
  vector<int> pr;
  bool prime[10000000];
void Sieve(ll int n) 
{ 
    memset(prime, true, sizeof(prime)); 
    prime[1]=false;
  
    for (int p=2; p*p<=n; p++) 
    { 
        if (prime[p] == true) 
        { 
            pr.push_back(p);
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
}
 


//It reurns Fn,Fn+1
pair<int, int> fib (int n) {
    if (n == 0)
        return {0, 1};

    auto p = fib(n >> 1);
    int c = p.first * (2 * p.second - p.first);
    int d = p.first * p.first + p.second * p.second;
    if (n & 1)
        return {d, c + d};
    else
        return {c, d};
}
int sum(int n)  
{  
    if (n == 0)  
    return 0;  
    return (n % 10 + sum(n / 10));  
}  
ll int factorial(ll int n,ll int m) 
{ 
    ll int fact = 1; 
    for(int i=2;i<=n;i++) { 
        fact = (fact * i)%m; 
    } 
    return fact%m; 
} 

int reduce(int x, int y) 
{ 
    int d; 
    d = __gcd(x, y); 
  
    x = x / d; 
    y = y / d; 
  
    return x;
} 
vector<ll int>v[100005];
ll int vis[100005];
ll int l=0;
void dfs(int n)
{
    vis[n]=1;
    l++;
    for(int i=0;i<v[n].size();i++)
    {
        if(vis[v[n][i]]==0)
        {
            dfs(v[n][i]);
           // l++;
        }
    }
}

int main() {
   ios;
   ll int n,k;
   cin>>n>>k;
   ll int p=n;
   n--;
   while(n--)
   {
      ll int x,y,z;
      cin>>x>>y>>z;
      if(z==0)
      {
          v[x].push_back(y);
          v[y].push_back(x);
      }
   }
   ll int s=0;
   for(int i=1;i<=p;i++)
   {
       if(vis[i]==0)
       {
           l=0;
           dfs(i);
           //cout<<l<<endl;
           //cout<<l<<endl;
           s=(s+power(l,k,mod));
       }
   }
   ll int o=power(p,k,mod);
   ll int ans=o-s;
   ans += mod;
   ans %= mod;
   cout<<ans<<endl;
}
