#include <bits/stdc++.h>
using namespace std;
int N, K;
long long m, rt, arr[200005], f[2][200005], g[2][200005], sz[200005],
    mx[200005], vis[200005], ans;
vector<int> edges[200005][2];
void add(int u, int fa, long long num, long long sum, int col, int fst) {
  sum += arr[u];
  f[fst][num] = (f[fst][num] + sum) % 1000000007;
  g[fst][num]++;
  ;
  m = max(m, num);
  for (int op : {0, 1}) {
    for (int v : edges[u][op]) {
      if (v != fa && !vis[v]) {
        add(v, u, num + (op != col), sum, op, fst);
      }
    }
  }
  return;
}
void upd(long long k, long long val) {
  for (int i = 1; i <= m; ++i) {
    g[0][i] += g[0][i - 1];
    g[1][i] += g[1][i - 1];
  }
  if (val != -1)
    ans += (g[0][min(K, (int)m)] + g[1][min(K, (int)m)]) * val * k % 1000000007;
  for (int op : {0, 1}) {
    for (int i = 1; i <= m; ++i) {
      ans += k * f[op][i] * g[op][min(K - i + 1, (int)m)] % 1000000007;
      ans += k * f[op][i] * g[op ^ 1][min(K - i, (int)m)] % 1000000007;
      if (i <= K && val != -1) ans += f[op][i] * k;
    }
  }
  ans %= 1000000007;
  for (int i = 1; i <= m; ++i) f[0][i] = f[1][i] = g[0][i] = g[1][i] = 0;
  return;
}
void calc(int u, int fa, int num) {
  mx[u] = 0;
  for (int op : {0, 1}) {
    for (int v : edges[u][op]) {
      if (!vis[v] && v != fa) {
        calc(v, u, num);
        mx[u] = max(mx[u], sz[v]);
      }
    }
  }
  mx[u] = max(mx[u], num - sz[u]);
  if (mx[u] < mx[rt]) rt = u;
  return;
}
void dfs(int u, int fa) {
  sz[u] = 1;
  for (int op : {0, 1}) {
    for (int v : edges[u][op]) {
      if (!vis[v] && v != fa) {
        dfs(v, u);
        sz[u] += sz[v];
      }
    }
  }
  return;
}
void solve(int x) {
  vis[x] = 1;
  m = 0;
  for (int op : {0, 1}) {
    for (int v : edges[x][op]) {
      if (!vis[v]) {
        add(v, 0, 1, arr[x] / 2, op, op);
      }
    }
  }
  upd(1, arr[x] / 2);
  for (int op : {0, 1}) {
    m = 0;
    for (int v : edges[x][op]) {
      if (!vis[v]) {
        add(v, 0, 1, arr[x] / 2, op, op);
        upd(-1, -1);
      }
    }
  }
  for (int op : {0, 1}) {
    for (int v : edges[x][op]) {
      if (!vis[v]) {
        rt = 0;
        dfs(v, 0);
        calc(v, 0, sz[v]);
        solve(rt);
      }
    }
  }
  return;
}
int main() {
  cin >> N >> K;
  K++;
  for (int i = 1; i <= N; ++i) {
    cin >> arr[i];
    arr[i] *= 2;
    ans += arr[i];
  }
  for (int i = 1; i < N; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    edges[a][c].push_back(b);
    edges[b][c].push_back(a);
  }
  mx[rt = 0] = N + 1;
  dfs(1, 0);
  calc(1, 0, sz[1]);
  solve(rt);
  ans += 1000000007;
  ans = ans * ((1000000007 + 1) / 2) % 1000000007;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
int N, K;
long long m, rt, arr[200005], f[2][200005], g[2][200005], sz[200005],
    mx[200005], vis[200005], ans;
vector<int> edges[200005][2];
void add(int u, int fa, long long num, long long sum, int col, int fst) {
  sum += arr[u];
  f[fst][num] = (f[fst][num] + sum) % 1000000007;
  g[fst][num]++;
  ;
  m = max(m, num);
  for (int op : {0, 1}) {
    for (int v : edges[u][op]) {
      if (v != fa && !vis[v]) {
        add(v, u, num + (op != col), sum, op, fst);
      }
    }
  }
  return;
}
void upd(long long k, long long val) {
  for (int i = 1; i <= m; ++i) {
    g[0][i] += g[0][i - 1];
    g[1][i] += g[1][i - 1];
  }
  if (val != -1)
    ans += (g[0][min(K, (int)m)] + g[1][min(K, (int)m)]) * val * k % 1000000007;
  for (int op : {0, 1}) {
    for (int i = 1; i <= m; ++i) {
      ans += k * f[op][i] * g[op][min(K - i + 1, (int)m)] % 1000000007;
      ans += k * f[op][i] * g[op ^ 1][min(K - i, (int)m)] % 1000000007;
      if (i <= K && val != -1) ans += f[op][i] * k;
    }
  }
  ans %= 1000000007;
  for (int i = 1; i <= m; ++i) f[0][i] = f[1][i] = g[0][i] = g[1][i] = 0;
  return;
}
void calc(int u, int fa, int num) {
  mx[u] = 0;
  for (int op : {0, 1}) {
    for (int v : edges[u][op]) {
      if (!vis[v] && v != fa) {
        calc(v, u, num);
        mx[u] = max(mx[u], sz[v]);
      }
    }
  }
  mx[u] = max(mx[u], num - sz[u]);
  if (mx[u] < mx[rt]) rt = u;
  return;
}
void dfs(int u, int fa) {
  sz[u] = 1;
  for (int op : {0, 1}) {
    for (int v : edges[u][op]) {
      if (!vis[v] && v != fa) {
        dfs(v, u);
        sz[u] += sz[v];
      }
    }
  }
  return;
}
void solve(int x) {
  vis[x] = 1;
  m = 0;
  for (int op : {0, 1}) {
    for (int v : edges[x][op]) {
      if (!vis[v]) {
        add(v, 0, 1, arr[x] / 2, op, op);
      }
    }
  }
  upd(1, arr[x] / 2);
  for (int op : {0, 1}) {
    m = 0;
    for (int v : edges[x][op]) {
      if (!vis[v]) {
        add(v, 0, 1, arr[x] / 2, op, op);
        upd(-1, -1);
      }
    }
  }
  for (int op : {0, 1}) {
    for (int v : edges[x][op]) {
      if (!vis[v]) {
        rt = 0;
        dfs(v, 0);
        calc(v, 0, sz[v]);
        solve(rt);
      }
    }
  }
  return;
}
int main() {
  cin >> N >> K;
  K++;
  for (int i = 1; i <= N; ++i) {
    cin >> arr[i];
    arr[i] *= 2;
    ans += arr[i];
  }
  for (int i = 1; i < N; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    edges[a][c].push_back(b);
    edges[b][c].push_back(a);
  }
  mx[rt = 0] = N + 1;
  dfs(1, 0);
  calc(1, 0, sz[1]);
  solve(rt);
  ans += 1000000007;
  ans = ans * ((1000000007 + 1) / 2) % 1000000007;
  cout << ans << endl;
  return 0;
}

```