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;
using ll = long long;
using ld = double;
bool DEBUGGING = 0;
void useiostream() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
}
void drop(char const *bu) {
  printf("%s", bu);
  exit(0);
}
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ld const pi = acos(-1.0);
ld const eps = 0.000000001;
template <typename T, typename TT>
bool mi(T &a, TT b) {
  return a > b ? (a = b, true) : false;
}
template <typename T, typename TT>
bool ma(T &a, TT b) {
  return a < b ? (a = b, true) : false;
}
mt19937 ran(time(0));
int const mod = (int)1e9 + 7;
int bp(int a, int p) {
  int re = 1;
  while (p) {
    if (p & 1) re = (re * 1LL * a) % mod;
    p >>= 1;
    a = (a * 1LL * a) % mod;
  }
  return re;
}
int inv(int a) { return bp(a, mod - 2); }
int n, k;
vector<int> g[100005];
set<int> black[100005];
int color[100005];
vector<int> ng[100005];
int cnt[100005];
int sz[100005];
void dfs(int ind, int col) {
  color[ind] = col;
  cnt[col]++;
  for (auto &x : g[ind]) {
    if (black[ind].count(x)) continue;
    if (color[x] == 0) dfs(x, col);
  }
}
int used[100005];
int res = 0;
int mul[100005];
int todel[100005];
int re[100005];
int get(int from, int what) { return bp(from, what); }
int dfsf(int ind) {
  sz[ind] = cnt[ind];
  for (auto &x : ng[ind]) {
    if (sz[x] == 0) {
      sz[ind] += dfsf(x);
    }
  }
  return sz[ind];
}
void dfsc(int ind) {
  used[ind] = 1;
  todel[ind] = bp(cnt[ind], k);
  for (auto &x : ng[ind]) {
    if (!used[x]) {
      dfsc(x);
      todel[ind] += todel[x];
      todel[ind] %= mod;
    }
  }
  re[ind] = (bp(sz[ind], k) - todel[ind] + mod) % mod;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
    if (c == 1) {
      black[a].insert(b);
      black[b].insert(a);
    }
  }
  int kek = 1;
  for (int i = 0; i < n; i++) {
    if (color[i] == 0) {
      dfs(i, kek++);
    }
  }
  for (int i = 0; i < n; i++) {
    for (auto &x : black[i]) {
      ng[color[x]].push_back(color[i]);
    }
  }
  dfsf(1);
  dfsc(1);
  printf("%d", re[1]);
  return 0;
}
