Problem: You are given a weighted tree consisting of n vertices. Recall that a tree is a connected graph without cycles. Vertices u_i and v_i are connected by an edge with weight w_i.

You are given m queries. The i-th query is given as an integer q_i. In this query you need to calculate the number of pairs of vertices (u, v) (u < v) such that the maximum weight of an edge on a simple path between u and v doesn't exceed q_i.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of vertices in the tree and the number of queries.

Each of the next n - 1 lines describes an edge of the tree. Edge i is denoted by three integers u_i, v_i and w_i — the labels of vertices it connects (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) and the weight of the edge (1 ≤ w_i ≤ 2 ⋅ 10^5). It is guaranteed that the given edges form a tree.

The last line of the input contains m integers q_1, q_2, ..., q_m (1 ≤ q_i ≤ 2 ⋅ 10^5), where q_i is the maximum weight of an edge in the i-th query.

Output

Print m integers — the answers to the queries. The i-th value should be equal to the number of pairs of vertices (u, v) (u < v) such that the maximum weight of an edge on a simple path between u and v doesn't exceed q_i.

Queries are numbered from 1 to m in the order of the input.

Examples

Input


7 5
1 2 1
3 2 3
2 4 1
4 5 2
5 7 4
3 6 2
5 2 3 4 1


Output


21 7 15 21 3 


Input


1 2
1 2


Output


0 0 


Input


3 3
1 2 1
2 3 2
1 3 2


Output


1 3 3 

Note

The picture shows the tree from the first example: <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
template <class T>
inline T read() {
  T x = 0, f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  };
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + s - 48;
    s = getchar();
  };
  return x * f;
}
template <typename T>
inline void wri(T x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  };
  if (x / 10) wri(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void print(T x, char s) {
  wri(x);
  putchar(s);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
const int maxn = 205;
int n, k, x, y;
vector<int> v[maxn];
int a[maxn];
int dp[maxn][maxn];
void dfs(int u, int fa) {
  int len = v[u].size();
  for (register int i = (0); i < (len); i++) {
    int y = v[u][i];
    if (y == fa) continue;
    dfs(y, u);
  }
  dp[u][0] = a[u];
  for (register int i = (0); i < (len); i++) {
    int y = v[u][i];
    if (y == fa) continue;
    dp[u][0] += dp[y][k];
  }
  for (register int i = (1); i < (n); i++) {
    for (register int j = (0); j < (len); j++) {
      int temp = 0;
      int y = v[u][j];
      if (y == fa) continue;
      temp += dp[y][i - 1];
      for (register int t = (0); t < (len); t++) {
        int p = v[u][t];
        if (p == fa || p == y) continue;
        temp += dp[p][max(i - 1, k - i)];
      }
      dp[u][i] = max(dp[u][i], temp);
    }
  }
  for (register int i = (n); i >= (0); i--)
    dp[u][i] = max(dp[u][i], dp[u][i + 1]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> k;
  memset(dp, 0, sizeof(dp));
  for (register int i = (1); i < (n + 1); i++) v[i].clear(), cin >> a[i];
  for (register int i = (1); i < (n); i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, -1);
  cout << dp[1][0] << endl;
  return 0;
}