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 mod = 1e9 + 7;
  template <typename T>
  struct fenwick {
    vector<T> a;
    int n;
    fenwick(int n) : n(n), a(n) {}
    void add(int x, T v) {
      assert(0 <= x && x < n);
      for (int i = x + 1; i <= n; i += i & -i) {
        a[i - 1] += v;
        if (a[i - 1] >= mod) a[i - 1] -= mod;
      }
    }
    T qry(int x) {
      T ret = 0;
      for (int i = min(x + 1, n); i > 0; i -= i & -i) ret += a[i - 1];
      return ret % mod;
    }
  };
  int main() {
    int n, k;
    cin >> n >> k;
    ++k;
    vector<vector<pair<int, int>>> g(n);
    vector<int> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    for (int i = 1; i < n; ++i) {
      int u, v, w;
      cin >> u >> v >> w;
      --u, --v;
      g[u].push_back({v, w});
      g[v].push_back({u, w});
    }
    int rt = -1, res = n;
    vector<int> mx(n), sz(n), vis(n);
    function<void(int, int)> dfs = [&](int u, int fa) {
      sz[u] = 1, mx[u] = 0;
      for (auto [v, w] : g[u]) {
        if (v == fa || vis[v]) continue;
        dfs(v, u);
        sz[u] += sz[v];
        mx[u] = max(mx[u], sz[v]);
      }
      mx[u] = max(mx[u], res - sz[u]);
      if (rt == -1 || mx[u] < mx[rt]) rt = u;
    };
    long long ans = 0;
    for (int i = 0; i < n; ++i) ans += a[i];
    function<void(int)> dfs2 = [&](int u) {
      int pre = res;
      vis[u] = 1;
      fenwick<long long> fen(res);
      fenwick<int> fen2(res);
      vector<vector<long long>> mp1(2, vector<long long>(res));
      vector<vector<int>> mp2(2, vector<int>(res));
      function<void(int, int, int, int, int, long long)> dfs3 =
          [&](int u, int fa, int pre, int cnt, int first, long long sum) {
            if (cnt > k) return;
            sum += a[u];
            if (sum >= mod) sum -= mod;
            mp1[first][cnt] += sum;
            if (mp1[first][cnt] >= mod) mp1[first][cnt] -= mod;
            mp2[first][cnt] += 1;
            fen.add(cnt, sum);
            fen2.add(cnt, 1);
            for (auto [v, w] : g[u]) {
              if (v == fa || vis[v]) continue;
              dfs3(v, u, w, cnt + (w != pre), first, sum);
            }
          };
      function<void(int, int, int, int, int, long long)> dfs4 =
          [&](int u, int fa, int pre, int cnt, int first, long long sum) {
            if (cnt > k) return;
            sum += a[u];
            if (sum >= mod) sum -= mod;
            ans += fen.qry(k - cnt) + fen2.qry(k - cnt) * sum % mod;
            if (k + 1 - cnt < res) {
              ans +=
                  mp1[first][k + 1 - cnt] + mp2[first][k + 1 - cnt] * sum % mod;
            }
            for (auto [v, w] : g[u]) {
              if (v == fa || vis[v]) continue;
              dfs4(v, u, w, cnt + (w != pre), first, sum);
            }
          };
      fen.add(0, a[u]);
      fen2.add(0, 1);
      for (auto [v, w] : g[u]) {
        if (vis[v]) continue;
        dfs4(v, u, w, 1, w, 0);
        dfs3(v, u, w, 1, w, a[u]);
      }
      for (auto [v, w] : g[u]) {
        if (vis[v]) continue;
        res = sz[v] < sz[u] ? sz[v] : pre - sz[u];
        rt = -1;
        dfs(v, u);
        dfs2(rt);
      }
      res = pre;
    };
    rt = -1;
    dfs(0, -1);
    dfs2(rt);
    cout << ans % mod << '\n';
  }
  