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>
using namespace std;
long long gcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int sum(long long a) {
  int sum = 0;
  while (a > 0) {
    sum = sum + (a % 10);
    a = a / 10;
  }
  return sum;
}
int count_digit(long long n) {
  int count = 0;
  while (n > 0) {
    n = n / 10;
    count++;
  }
  return count;
}
int binarySearch(int x, int y, long long z, long long v[]) {
  int low = x;
  int high = y;
  int mid = x + (y - x) / 2;
  while (low <= high) {
    if (v[mid] == z) return mid;
    if (v[mid] < z) return binarySearch(mid + 1, high, z, v);
    if (v[mid] > z) return binarySearch(low, mid - 1, z, v);
  }
  return -1;
}
long long modularExponentiation(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  else
    return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
long long binaryExponentiation(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
int binary(int n) {
  int c = 0;
  while (n > 0) {
    if (n % 2 == 1) {
      return pow(2, c);
    }
    n = n / 2;
    c++;
  }
}
set<long long> s;
void genrate(long long n, int len, int max) {
  if (len > max) return;
  s.insert(n);
  genrate(n * 10 + 1, len + 1, max);
  genrate(n * 10 + 0, len + 1, max);
}
vector<long long> ans;
vector<pair<long long, long long> > v[300005];
bool visited[300005];
int c = 0;
int dfs(int x) {
  visited[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i].second == 0 && visited[v[x][i].first] == 0) {
      c++;
      dfs(v[x][i].first);
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tests = 1;
  while (tests--) {
    long long n, k;
    cin >> n >> k;
    for (int i = 1; i < n; i++) {
      int x, y, z;
      cin >> x >> y >> z;
      v[x].push_back({y, z});
      v[y].push_back({x, z});
    }
    memset(visited, 0, sizeof(visited));
    for (int i = 1; i <= n; i++) {
      if (!visited[i]) {
        c = 1;
        dfs(i);
        ans.push_back(c);
      }
    }
    long long ans1 = modularExponentiation(n, k, 1000000007);
    for (int i = 0; i < ans.size(); i++) {
      ans1 = ans1 - modularExponentiation(ans[i], k, 1000000007);
    }
    cout << ans1;
  }
}
