#include <bits/stdc++.h>
using namespace std;
const int maxn = 410000;
const int mod = 1e9 + 7;
struct Tree {
  long long C[maxn];
  int n;
  void init(int N) {
    this->n = N + 4;
    memset(C, 0, sizeof(C[0]) * (n + 4));
  }
  void add(int x, long long d) {
    x += 1;
    for (int i = x; i <= n; i += ((i) & (-i))) C[i] = (C[i] + d) % mod;
  }
  long long sum(int x) {
    if (x < 0) return 0;
    x += 1;
    x = min(x, n);
    long long ret = 0;
    for (int i = x; i > 0; i -= ((i) & (-i))) ret += C[i];
    return ret % mod;
  }
} value[2], times[2];
vector<pair<int, int>> G[maxn];
int vis[maxn], siz[maxn], f[maxn], pos, sum;
int n, k, a[maxn];
long long answer = 0;
void getroot(int x, int fa) {
  f[x] = 0;
  siz[x] = 1;
  for (auto [y, w] : G[x])
    if (y != fa && !vis[y]) {
      getroot(y, x);
      f[x] = max(f[x], siz[y]);
      siz[x] += siz[y];
    }
  f[x] = max(f[x], sum - siz[x]);
  if (f[x] < f[pos]) pos = x;
}
void dfs1(int x, int fa, long long w, int color, int cnt, int start) {
  w = (w + a[x]) % mod;
  answer += value[start].sum(k - cnt) % mod;
  answer += times[start].sum(k - cnt) * w % mod;
  answer += value[start ^ 1].sum(k - cnt - 1) % mod;
  answer += times[start ^ 1].sum(k - cnt - 1) * w % mod;
  for (auto [y, t] : G[x])
    if (y != fa && !vis[y]) {
      dfs1(y, x, w, t, cnt + (t != color), start);
    }
}
void dfs2(int x, int fa, long long w, int color, int cnt, int start) {
  w = (w + a[x]) % mod;
  if (cnt <= k) {
    answer += w;
    value[start].add(cnt, w);
    times[start].add(cnt, 1);
  }
  for (auto [y, t] : G[x])
    if (y != fa && !vis[y]) {
      dfs2(y, x, w, t, cnt + (t != color), start);
    }
}
int travel(int x, int fa) {
  int ret = 1;
  for (auto [y, t] : G[x])
    if (y != fa && !vis[y]) {
      ret += travel(y, x);
    }
  return ret;
}
void calc(int x) {
  answer += a[x];
  const int size = travel(x, -1) + 1;
  for (int i = 0; i < 2; ++i) {
    value[i].init(size);
    times[i].init(size);
  }
  for (auto [y, t] : G[x])
    if (!vis[y]) {
      dfs1(y, x, 0, t, 0, t);
      dfs2(y, x, a[x], t, 0, t);
    }
  answer %= mod;
}
void solve(int x, int cnt) {
  pos = maxn - 1;
  f[pos] = sum = cnt;
  getroot(x, -1);
  int root = pos;
  vis[root] = 1;
  calc(root);
  for (auto [y, w] : G[root])
    if (!vis[y]) solve(y, siz[y] < siz[root] ? siz[y] : cnt - siz[root]);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i < n; ++i) {
    int x, y, w;
    scanf("%d %d %d", &x, &y, &w);
    G[x].emplace_back(y, w);
    G[y].emplace_back(x, w);
  }
  solve(1, n);
  answer %= mod;
  printf("%lld\n", answer);
  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;
const int maxn = 410000;
const int mod = 1e9 + 7;
struct Tree {
  long long C[maxn];
  int n;
  void init(int N) {
    this->n = N + 4;
    memset(C, 0, sizeof(C[0]) * (n + 4));
  }
  void add(int x, long long d) {
    x += 1;
    for (int i = x; i <= n; i += ((i) & (-i))) C[i] = (C[i] + d) % mod;
  }
  long long sum(int x) {
    if (x < 0) return 0;
    x += 1;
    x = min(x, n);
    long long ret = 0;
    for (int i = x; i > 0; i -= ((i) & (-i))) ret += C[i];
    return ret % mod;
  }
} value[2], times[2];
vector<pair<int, int>> G[maxn];
int vis[maxn], siz[maxn], f[maxn], pos, sum;
int n, k, a[maxn];
long long answer = 0;
void getroot(int x, int fa) {
  f[x] = 0;
  siz[x] = 1;
  for (auto [y, w] : G[x])
    if (y != fa && !vis[y]) {
      getroot(y, x);
      f[x] = max(f[x], siz[y]);
      siz[x] += siz[y];
    }
  f[x] = max(f[x], sum - siz[x]);
  if (f[x] < f[pos]) pos = x;
}
void dfs1(int x, int fa, long long w, int color, int cnt, int start) {
  w = (w + a[x]) % mod;
  answer += value[start].sum(k - cnt) % mod;
  answer += times[start].sum(k - cnt) * w % mod;
  answer += value[start ^ 1].sum(k - cnt - 1) % mod;
  answer += times[start ^ 1].sum(k - cnt - 1) * w % mod;
  for (auto [y, t] : G[x])
    if (y != fa && !vis[y]) {
      dfs1(y, x, w, t, cnt + (t != color), start);
    }
}
void dfs2(int x, int fa, long long w, int color, int cnt, int start) {
  w = (w + a[x]) % mod;
  if (cnt <= k) {
    answer += w;
    value[start].add(cnt, w);
    times[start].add(cnt, 1);
  }
  for (auto [y, t] : G[x])
    if (y != fa && !vis[y]) {
      dfs2(y, x, w, t, cnt + (t != color), start);
    }
}
int travel(int x, int fa) {
  int ret = 1;
  for (auto [y, t] : G[x])
    if (y != fa && !vis[y]) {
      ret += travel(y, x);
    }
  return ret;
}
void calc(int x) {
  answer += a[x];
  const int size = travel(x, -1) + 1;
  for (int i = 0; i < 2; ++i) {
    value[i].init(size);
    times[i].init(size);
  }
  for (auto [y, t] : G[x])
    if (!vis[y]) {
      dfs1(y, x, 0, t, 0, t);
      dfs2(y, x, a[x], t, 0, t);
    }
  answer %= mod;
}
void solve(int x, int cnt) {
  pos = maxn - 1;
  f[pos] = sum = cnt;
  getroot(x, -1);
  int root = pos;
  vis[root] = 1;
  calc(root);
  for (auto [y, w] : G[root])
    if (!vis[y]) solve(y, siz[y] < siz[root] ? siz[y] : cnt - siz[root]);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i < n; ++i) {
    int x, y, w;
    scanf("%d %d %d", &x, &y, &w);
    G[x].emplace_back(y, w);
    G[y].emplace_back(x, w);
  }
  solve(1, n);
  answer %= mod;
  printf("%lld\n", answer);
  return 0;
}

```