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>
#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int MAXN = 1e5 + 1;
int par[MAXN];
int ran[MAXN];
void init(int n) {
  for (long long i = 0; i < n; ++i) {
    par[i] = i;
    ran[i] = 0;
  }
}
int find(int x) {
  if (par[x] == x) {
    return x;
  } else {
    return par[x] = find(par[x]);
  }
}
void unite(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  if (ran[x] < ran[y]) {
    par[x] = y;
  } else {
    par[y] = x;
    if (ran[x] == ran[y]) ran[x]++;
  }
}
bool same(int x, int y) { return find(x) == find(y); }
long long quickpow(long long x, long long n) {
  long long ans = 1ll;
  while (n) {
    if (n & 1) {
      ans *= x;
      ans %= mod;
    }
    x *= x;
    x %= mod;
    n >>= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  init(n);
  for (long long i = 0; i < n - 1; ++i) {
    int x, y, c;
    cin >> x >> y >> c;
    if (c == 0) {
      unite(x - 1, y - 1);
    }
  }
  map<int, long long> cnt;
  for (long long i = 0; i < n; ++i) {
    cnt[find(i)]++;
  }
  long long ans = 0;
  for (auto e : cnt) {
    long long x = e.second;
    if (x > 1) {
      ans += quickpow(x, k);
      ans %= mod;
    } else {
      ans++;
      ans %= mod;
    }
  }
  ans = quickpow(n, k) - ans;
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans << endl;
}
