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;
string alpha = "abcdefghijklmnopqrstuvwxyz";
int MOD = 1000000007;
const int N = 100100;
double Len(long long x, long long y, long long X, long long Y) {
  return sqrt((X - x) * (X - x) + (Y - y) * (Y - y));
}
int n, k;
int dp[105][N][2];
int cost[N];
vector<pair<int, int> > adj[N];
int const nN = 2 * N;
int const logN = 20;
int id, ID[N], rID[N], First[N], ST[nN][logN], lg[nN];
vector<int> vec;
void dfs(int cur, int par) {
  ID[cur] = id++;
  rID[ID[cur]] = cur;
  vec.push_back(ID[cur]);
  for (auto x : adj[cur])
    if (x.first != par)
      cost[x.first] = cost[cur] + x.second, dfs(x.first, cur),
      vec.push_back(ID[cur]);
}
inline int LCA(int a, int b) {
  int L, R, idx, d, lca;
  a = ID[a];
  b = ID[b];
  L = First[a], R = First[b];
  if (L > R) swap(L, R);
  d = R - L + 1;
  d = lg[d];
  idx = R - (1 << d) + 1;
  lca = min(ST[L][d], ST[idx][d]);
  lca = rID[lca];
  a = rID[a];
  b = rID[b];
  return lca;
}
inline void Build() {
  int cnt = 0, two = 1;
  while (two < nN) lg[two] = cnt++, two *= 2;
  for (int i = 3; i < nN; ++i)
    if (!lg[i]) lg[i] = lg[i - 1];
  int nN = (vec.size());
  memset(First, -1, sizeof First);
  for (int i = 0; i < nN; ++i)
    ST[i][0] = vec[i], (First[vec[i]] == -1 ? First[vec[i]] = i : 0);
  for (int i = 1; (1 << i) <= nN; ++i)
    for (int j = 0; j + (1 << i) - 1 < nN; ++j)
      ST[j][i] = min(ST[j][i - 1], ST[j + (1 << (i - 1))][i - 1]);
}
int solve(int idx, int last, bool ok) {
  if (idx == k) return ok;
  int& ret = dp[idx][last][ok];
  if (~ret) return ret;
  ret = 0;
  for (int i = 1; i <= n; ++i) {
    int x = i;
    int y = last;
    int lca = LCA(x, y);
    int nok = ok;
    if (lca == x || lca == y)
      nok = ok || (abs(cost[x] - cost[y]) > 0);
    else
      nok = ok || (abs(cost[x] + cost[y]) > 0);
    ret += solve(idx + 1, i, nok);
    ret %= MOD;
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = 0; i < n - 1; ++i) {
    int x, y, c;
    scanf("%d", &x);
    scanf("%d", &y);
    scanf("%d", &c);
    adj[x].push_back({y, c});
    adj[y].push_back({x, c});
  }
  dfs(1, 0);
  Build();
  memset(dp, -1, sizeof(dp));
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += solve(1, i, 0);
    ans %= MOD;
  }
  cout << ans << "\n";
  return 0;
}
