description: There is a city park represented as a tree with n attractions as its vertices and n - 1 rails as its edges. The i-th attraction has happiness value a_i.
  
  Each rail has a color. It is either black if t_i = 0, or white if t_i = 1. Black trains only operate on a black rail track, and white trains only operate on a white rail track. If you are previously on a black train and want to ride a white train, or you are previously on a white train and want to ride a black train, you need to use 1 ticket.
  
  The path of a tour must be a simple path — it must not visit an attraction more than once. You do not need a ticket the first time you board a train. You only have k tickets, meaning you can only switch train types at most k times. In particular, you do not need a ticket to go through a path consisting of one rail color.
  
  Define f(u, v) as the sum of happiness values of the attractions in the tour (u, v), which is a simple path that starts at the u-th attraction and ends at the v-th attraction. Find the sum of f(u,v) for all valid tours (u, v) (1 ≤ u ≤ v ≤ n) that does not need more than k tickets, modulo 10^9 + 7.
  
  Input
  
  The first line contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ k ≤ n-1) — the number of attractions in the city park and the number of tickets you have.
  
  The second line contains n integers a_1, a_2,…, a_n (0 ≤ a_i ≤ 10^9) — the happiness value of each attraction.
  
  The i-th of the next n - 1 lines contains three integers u_i, v_i, and t_i (1 ≤ u_i, v_i ≤ n, 0 ≤ t_i ≤ 1) — an edge between vertices u_i and v_i with color t_i. The given edges form a tree.
  
  Output
  
  Output an integer denoting the total happiness value for all valid tours (u, v) (1 ≤ u ≤ v ≤ n), modulo 10^9 + 7.
  
  Examples
  
  Input
  
  
  5 0
  1 3 2 6 4
  1 2 1
  1 4 0
  3 2 1
  2 5 0
  
  
  Output
  
  
  45
  
  
  Input
  
  
  3 1
  1 1 1
  1 2 1
  3 2 0
  
  
  Output
  
  
  10
solution: #include <bits/stdc++.h>
  using namespace std;
  const int p = 1e9 + 7;
  const int N = 2e5 + 10;
  int n, k, a[N];
  vector<pair<int, int> > g[N];
  inline void A(int &x, int y) { x = (x + y >= p ? x + y - p : x + y); }
  inline void U(int &x, int y) { x = (x > y ? x : y); }
  int del[N], si[N];
  int mnsi, mnid;
  inline void GetSi(int x, int fa) {
    si[x] = 1;
    for (auto e : g[x])
      if (e.second != fa && !del[e.second]) {
        GetSi(e.second, x);
        si[x] += si[e.second];
      }
  }
  inline void GetMx(int x, int fa, int al) {
    int mx = al - si[x];
    for (auto e : g[x])
      if (e.second != fa && !del[e.second]) {
        mx = max(mx, si[e.second]);
        GetMx(e.second, x, al);
      }
    if (mx < mnsi) mnsi = mx, mnid = x;
  }
  struct DS {
    int f[N];
    inline void Add(int x, int y) {
      for (int i = x; i <= n; i += (i & -i)) A(f[i], y);
    }
    inline void Clr(int x) {
      for (int i = x; i <= n; i += (i & -i)) f[i] = 0;
    }
    inline int Ask(int x) {
      int ans = 0;
      for (int i = x; i; i -= (i & -i)) A(ans, f[i]);
      return ans;
    }
  } f1, g1, f2, g2;
  int sw1, sw2;
  int ans;
  void PushIn(int x, int fa, int fe, int sm, int sw, DS &f, DS &G, int &mxsw) {
    if (sw > k) return;
    mxsw = max(mxsw, sw);
    A(sm, a[x]);
    f.Add(sw + 1, sm);
    G.Add(sw + 1, 1);
    for (auto e : g[x]) {
      int y = e.second, z = e.first;
      if (del[y] || y == fa) continue;
      PushIn(y, x, z, sm, sw + (fe != z), f, G, mxsw);
    }
  }
  void Calc(int x, int fa, int fe, int sm, int sw) {
    if (sw > k) return;
    A(sm, a[x]);
    if (sw < k)
      ans = (ans + 1ll * sm * (g1.Ask(k - sw) - g2.Ask(k - sw)) +
             (f1.Ask(k - sw) - f2.Ask(k - sw))) %
            p;
    ans = (ans + 1ll * sm * g2.Ask(k - sw + 1) + f2.Ask(k - sw + 1)) % p;
    for (auto e : g[x]) {
      int y = e.second, z = e.first;
      if (del[y] || y == fa) continue;
      Calc(y, x, z, sm, sw + (fe != z));
    }
  }
  inline void Solve(int x) {
    GetSi(x, 0);
    mnsi = 1e9;
    GetMx(x, 0, si[x]);
    int u = mnid;
    sw1 = 0, sw2 = 0;
    f1.Add(1, a[u]);
    f2.Add(1, a[u]);
    g1.Add(1, 1);
    g2.Add(1, 1);
    A(ans, a[u]);
    int la = -1;
    for (auto e : g[u]) {
      int y = e.second, z = e.first;
      if (z != la) {
        for (int i = 1; i <= sw2 + 1; ++i) f2.Clr(i), g2.Clr(i);
        sw2 = 0;
        la = z;
        f2.Add(1, a[u]);
        g2.Add(1, 1);
      }
      if (del[y]) continue;
      Calc(y, u, z, 0, 0);
      PushIn(y, u, z, a[u], 0, f1, g1, sw1);
      PushIn(y, u, z, a[u], 0, f2, g2, sw2);
    }
    for (int i = 1; i <= sw1 + 1; ++i) f1.Clr(i), g1.Clr(i);
    for (int i = 1; i <= sw2 + 1; ++i) f2.Clr(i), g2.Clr(i);
    del[u] = 1;
    for (auto e : g[u])
      if (!del[e.second]) Solve(e.second);
  }
  inline void Dfs(int x, int fa, int fe, int sw, int sm, int rt) {
    if (sw > k) return;
    A(sm, a[x]);
    if (x <= rt) A(ans, p - sm);
    for (auto e : g[x]) {
      int y = e.second, z = e.first;
      if (y == fa) continue;
      Dfs(y, x, z, sw + (fe != z), sm, rt);
    }
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%d", a + i);
    for (int i = 1, x, y, z; i < n; ++i) {
      scanf("%d%d%d", &x, &y, &z);
      g[x].push_back(pair<int, int>(z, y));
      g[y].push_back(pair<int, int>(z, x));
    }
    for (int i = 1; i <= n; ++i) sort(g[i].begin(), g[i].end());
    Solve(1);
    if (ans < 0) ans += p;
    printf("%d\n", ans);
    return 0;
  }
  