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>
  #pragma warning(disable : 4996)
  #pragma comment(linker, "/STACK:16777216")
  using namespace std;
  const int INF = 1000000000 + 1e8;
  const long long LINF = 2000000000000000000;
  const long long mod = 1e9 + 7;
  int n, k;
  vector<int> a;
  vector<vector<pair<int, int>>> g;
  vector<int> sz;
  vector<int> mark;
  struct BIT {
    int n;
    vector<long long> t;
    void init(int _n) {
      n = _n;
      t.resize(n);
    }
    void inc(int i, long long val) {
      for (; i < n; i = (i | (i + 1))) t[i] = (t[i] + val) % mod;
    }
    long long pref(int r) {
      long long res = 0;
      for (; r >= 0; r = (r & (r + 1)) - 1) res = (res + t[r]) % mod;
      return (res + mod) % mod;
    }
  };
  BIT pref0, pref1, cnt0, cnt1;
  void find_sz(int v, int p = -1) {
    sz[v] = 1;
    for (auto [to, type] : g[v])
      if (to != p && !mark[to]) {
        find_sz(to, v);
        sz[v] += sz[to];
      }
  }
  int find_centroid(int v, int p, int n) {
    for (auto [to, type] : g[v])
      if (!mark[to] && to != p && sz[to] * 2 > n) return find_centroid(to, v, n);
    return v;
  }
  struct history {
    int lastEdge;
    long long value;
    int k;
  };
  vector<history> current;
  vector<history> total;
  long long ans = 0;
  void go(int v, int p, history hist, int last, int c) {
    hist.value = (hist.value + a[v]) % mod;
    if (hist.k <= k) {
      ans = (ans + hist.value) % mod;
      int i = k - hist.k;
      ans = (ans + pref0.pref(i - (hist.lastEdge == 1))) % mod;
      ans = (ans +
             cnt0.pref(i - (hist.lastEdge == 1)) * (hist.value - a[c] + mod)) %
            mod;
      ans = (ans + pref1.pref(i - (hist.lastEdge == 0)));
      ans = (ans +
             cnt1.pref(i - (hist.lastEdge == 0)) * (hist.value - a[c] + mod)) %
            mod;
      current.push_back(hist);
    }
    for (auto [to, type] : g[v]) {
      if (!mark[to] && to != p) {
        hist.k += (type != last);
        go(to, v, hist, type, c);
        hist.k -= (type != last);
      }
    }
  }
  void calc(int v) {
    find_sz(v);
    int c = find_centroid(v, -1, sz[v]);
    mark[c] = true;
    ans = (ans + a[c]) % mod;
    for (auto [to, type] : g[c]) {
      if (!mark[to]) {
        go(to, c, {type, a[c], 0}, type, c);
        while (!current.empty()) {
          if (type == 0) {
            pref0.inc(current.back().k, current.back().value);
            cnt0.inc(current.back().k, 1);
          } else {
            pref1.inc(current.back().k, current.back().value);
            cnt1.inc(current.back().k, 1);
          }
          total.push_back(current.back());
          current.pop_back();
        }
      }
    }
    while (!total.empty()) {
      if (total.back().lastEdge == 0) {
        pref0.inc(total.back().k, -total.back().value);
        cnt0.inc(total.back().k, -1);
      } else {
        pref1.inc(total.back().k, -total.back().value);
        cnt1.inc(total.back().k, -1);
      }
      total.pop_back();
    }
    for (auto [to, type] : g[c]) {
      if (!mark[to]) {
        calc(to);
      }
    }
  }
  void solve() {
    cin >> n >> k;
    a.resize(n);
    g.resize(n);
    mark.resize(n);
    sz.resize(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n - 1; i++) {
      int u, v, t;
      cin >> u >> v >> t;
      u--, v--;
      g[u].push_back({v, t});
      g[v].push_back({u, t});
    }
    pref0.init(n);
    pref1.init(n);
    cnt0.init(n);
    cnt1.init(n);
    current.reserve(n);
    total.reserve(n);
    calc(0);
    cout << ans;
  }
  int main() {
    srand(time(0));
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tst = 1;
    while (tst--) {
      solve();
    }
  }
  