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/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#include <ext/pb_ds/detail/standard_policies.hpp>
 
using namespace __gnu_pbds; 
using namespace std;
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define eb emplace_back
#define F first 
#define S second
#define minHeap priority_queue<int,vector<int>,greater<int>>
#define MAX(a,b,c) max(a,max(b,c))
typedef pair<int,int> pi;
typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag,tree_order_statistics_node_update> pbds; 
 
const char nl ='\n';const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 7;
 
const int N=1e6+1; 
const double pie= 3.1415926535;
 
int dxx[]={-1,1,0,0,1,-1,-1,1};
int dyy[]={0,0,-1,1,1,-1,1,-1};
int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1}; 
 
void IO() { 
        #ifndef ONLINE_JUDGE    
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout); 
        #endif
   }

ll getpower(ll n, ll k)
{
       ll ans=1;
       while(k>0)
       {
              if(k&1)
              {
                ans=(n*ans)%MOD;
              }
              n=(n*n)%MOD;
              k/=2;

       }
       return ans;
}




void decode()
{       
              int n,k;
              cin>>n>>k;
              int white=0;
              for(int i=0; i<n-1; i++)
              {
                    int x,y,z;
                    cin>>x>>y>>z;
                    white+=(z==0);
              }
              ll total=getpower(n,k);
         
              ll getwhite=getpower(white+1,k);
              ll ans=total-getwhite-(n-white-1);
              cout<<ans<<endl;    
}
 
 
 
int32_t main()
{
         IO(); 
         IOS;
         int T=1;  
        // cin>>T;
         while(T--)  
          {
                decode();   
              
          }
        return 0;
}