#include <bits/stdc++.h>
using namespace std;
long long res;
int N, K;
int A[200201];
vector<pair<int, int> > g[200201];
int sz[200201];
bool vis[200201];
long long fenC[200201], fenH[200201], fenCC[200201], fenHH[200201];
long long Sum(long long fen[], int idx) {
  idx++;
  idx = min(idx, N);
  if (idx < 0) return 0LL;
  long long ret = 0;
  while (idx) {
    ret += fen[idx], ret %= 1000000007;
    idx &= idx - 1;
  }
  return ret;
}
void Add(long long fen[], int idx, long long val) {
  idx++;
  while (idx <= N) {
    fen[idx] += val, fen[idx] %= 1000000007;
    idx += idx & -idx;
  }
}
void Del(long long fen[], int idx) {
  idx++;
  while (idx <= N) {
    fen[idx] = 0;
    idx += idx & -idx;
  }
}
void DFS(int n, int prev) {
  sz[n] = 1;
  for (auto [next, cost] : g[n]) {
    if (next == prev || vis[next]) continue;
    DFS(next, n);
    sz[n] += sz[next];
  }
}
int Centroid(int n, int prev, const int V) {
  for (auto [next, cost] : g[n]) {
    if (next == prev || vis[next]) continue;
    if (sz[next] * 2 > V) return Centroid(next, n, V);
  }
  return n;
}
vector<pair<int, long long> > v;
void DFS2(int n, int prev, int last, int d, long long h, int root) {
  if (d <= K) {
    res += h + A[root], res %= 1000000007;
  }
  v.push_back({d, h % 1000000007});
  for (auto [next, cost] : g[n]) {
    if (next == prev || vis[next]) continue;
    DFS2(next, n, cost, d + (cost != last), h + A[next], root);
  }
}
void dnc(int n) {
  vis[n] = true;
  res += A[n], res %= 1000000007;
  vector<int> del;
  for (auto [next, cost] : g[n]) {
    if (vis[next]) continue;
    if (cost == 1) continue;
    v.clear();
    DFS2(next, n, cost, 0, A[next], n);
    for (auto [d, h] : v) {
      long long cnt = Sum(fenC, K - d), val = Sum(fenH, K - d);
      res += (h + A[n]) * cnt + val, res %= 1000000007;
    }
    for (auto [d, h] : v) {
      Add(fenC, d, 1), Add(fenH, d, h);
      del.push_back(d);
    }
  }
  for (auto [next, cost] : g[n]) {
    if (vis[next]) continue;
    if (cost == 0) continue;
    v.clear();
    DFS2(next, n, cost, 0, A[next], n);
    for (auto [d, h] : v) {
      long long cnt = Sum(fenCC, K - d), val = Sum(fenHH, K - d);
      res += (h + A[n]) * cnt + val, res %= 1000000007;
      cnt = Sum(fenC, K - d - 1), val = Sum(fenH, K - d - 1);
      res += (h + A[n]) * cnt + val, res %= 1000000007;
    }
    for (auto [d, h] : v) {
      Add(fenCC, d, 1), Add(fenHH, d, h);
      del.push_back(d);
    }
  }
  for (int d : del) {
    Del(fenC, d);
    Del(fenH, d);
    Del(fenCC, d);
    Del(fenHH, d);
  }
  for (auto [next, cost] : g[n]) {
    if (vis[next]) continue;
    DFS(next, n);
    dnc(Centroid(next, n, sz[next]));
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(ios::badbit | ios::failbit);
  cin >> N >> K;
  for (int i = 1; i <= N; i++) cin >> A[i];
  for (int i = 0; i < N - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    g[a].push_back({b, c});
    g[b].push_back({a, c});
  }
  DFS(1, 0);
  dnc(Centroid(1, 0, N));
  res %= 1000000007;
  res += 1000000007;
  res %= 1000000007;
  cout << res << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long res;
int N, K;
int A[200201];
vector<pair<int, int> > g[200201];
int sz[200201];
bool vis[200201];
long long fenC[200201], fenH[200201], fenCC[200201], fenHH[200201];
long long Sum(long long fen[], int idx) {
  idx++;
  idx = min(idx, N);
  if (idx < 0) return 0LL;
  long long ret = 0;
  while (idx) {
    ret += fen[idx], ret %= 1000000007;
    idx &= idx - 1;
  }
  return ret;
}
void Add(long long fen[], int idx, long long val) {
  idx++;
  while (idx <= N) {
    fen[idx] += val, fen[idx] %= 1000000007;
    idx += idx & -idx;
  }
}
void Del(long long fen[], int idx) {
  idx++;
  while (idx <= N) {
    fen[idx] = 0;
    idx += idx & -idx;
  }
}
void DFS(int n, int prev) {
  sz[n] = 1;
  for (auto [next, cost] : g[n]) {
    if (next == prev || vis[next]) continue;
    DFS(next, n);
    sz[n] += sz[next];
  }
}
int Centroid(int n, int prev, const int V) {
  for (auto [next, cost] : g[n]) {
    if (next == prev || vis[next]) continue;
    if (sz[next] * 2 > V) return Centroid(next, n, V);
  }
  return n;
}
vector<pair<int, long long> > v;
void DFS2(int n, int prev, int last, int d, long long h, int root) {
  if (d <= K) {
    res += h + A[root], res %= 1000000007;
  }
  v.push_back({d, h % 1000000007});
  for (auto [next, cost] : g[n]) {
    if (next == prev || vis[next]) continue;
    DFS2(next, n, cost, d + (cost != last), h + A[next], root);
  }
}
void dnc(int n) {
  vis[n] = true;
  res += A[n], res %= 1000000007;
  vector<int> del;
  for (auto [next, cost] : g[n]) {
    if (vis[next]) continue;
    if (cost == 1) continue;
    v.clear();
    DFS2(next, n, cost, 0, A[next], n);
    for (auto [d, h] : v) {
      long long cnt = Sum(fenC, K - d), val = Sum(fenH, K - d);
      res += (h + A[n]) * cnt + val, res %= 1000000007;
    }
    for (auto [d, h] : v) {
      Add(fenC, d, 1), Add(fenH, d, h);
      del.push_back(d);
    }
  }
  for (auto [next, cost] : g[n]) {
    if (vis[next]) continue;
    if (cost == 0) continue;
    v.clear();
    DFS2(next, n, cost, 0, A[next], n);
    for (auto [d, h] : v) {
      long long cnt = Sum(fenCC, K - d), val = Sum(fenHH, K - d);
      res += (h + A[n]) * cnt + val, res %= 1000000007;
      cnt = Sum(fenC, K - d - 1), val = Sum(fenH, K - d - 1);
      res += (h + A[n]) * cnt + val, res %= 1000000007;
    }
    for (auto [d, h] : v) {
      Add(fenCC, d, 1), Add(fenHH, d, h);
      del.push_back(d);
    }
  }
  for (int d : del) {
    Del(fenC, d);
    Del(fenH, d);
    Del(fenCC, d);
    Del(fenHH, d);
  }
  for (auto [next, cost] : g[n]) {
    if (vis[next]) continue;
    DFS(next, n);
    dnc(Centroid(next, n, sz[next]));
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(ios::badbit | ios::failbit);
  cin >> N >> K;
  for (int i = 1; i <= N; i++) cin >> A[i];
  for (int i = 0; i < N - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    g[a].push_back({b, c});
    g[b].push_back({a, c});
  }
  DFS(1, 0);
  dnc(Centroid(1, 0, N));
  res %= 1000000007;
  res += 1000000007;
  res %= 1000000007;
  cout << res << "\n";
  return 0;
}

```