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;
inline int read() {
  int f = 1, x = 0;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * x;
}
const int MAXN = 100000 + 10;
int n, k;
vector<int> G[MAXN];
int a[MAXN];
int id[MAXN];
int dep[MAXN];
inline void dfs(int x, int fa) {
  for (int i = 0; i < G[x].size(); i++) {
    int v = G[x][i];
    if (v == fa) continue;
    dep[v] = dep[x] + 1;
    dfs(v, x);
  }
}
inline bool cmp(int a, int b) { return dep[a] > dep[b]; }
int dis[MAXN], q[MAXN];
inline int calc(int x) {
  memset(dis, -1, sizeof(dis));
  int h = 1, t = 0;
  q[++t] = x;
  dis[x] = 0;
  int res = a[x];
  while (h <= t) {
    int u = q[h];
    h++;
    a[u] -= res;
    if (dis[u] == k) continue;
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (dis[v] == -1) {
        t++;
        q[t] = v;
        dis[v] = dis[u] + 1;
      }
    }
  }
  return res;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read(), id[i] = i;
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  sort(id + 1, id + n + 1, cmp);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[id[i]] > 0) {
      ans += calc(id[i]);
    }
  }
  cout << ans << endl;
  return 0;
}