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 int maxn = 205;
int a[maxn];
vector<int> g[maxn];
int dp[maxn][maxn];
int n, k;
void dfs(int cur, int fa) {
  dp[cur][0] = a[cur];
  for (int i = 0; i < g[cur].size(); i++) {
    if (g[cur][i] == fa) continue;
    dfs(g[cur][i], cur);
  }
  for (int dep = 0; dep <= n; dep++) {
    if (dep == 0) {
      for (int i = 0; i < g[cur].size(); i++) {
        int to = g[cur][i];
        if (to == fa) continue;
        dp[cur][dep] += dp[to][max(0, k - dep - 1)];
      }
    } else {
      int sum = 0;
      for (int j = 0; j < g[cur].size(); j++) {
        int to = g[cur][j];
        if (to == fa) continue;
        sum += dp[to][max(dep - 1, k - dep - 1)];
      }
      for (int i = 0; i < g[cur].size(); i++) {
        int to = g[cur][i];
        if (to == fa) continue;
        int tmp = sum + dp[to][dep - 1] - dp[to][max(dep - 1, k - dep - 1)];
        dp[cur][dep] = max(dp[cur][dep], tmp);
      }
    }
  }
  for (int i = n - 2; i >= 0; i--) {
    dp[cur][i] = max(dp[cur][i], dp[cur][i + 1]);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  k++;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1);
  printf("%d\n", dp[0][0]);
}