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 dp[maxn][maxn];
int a[maxn];
int head[maxn];
int cnt;
int n, k;
struct edge {
  int to, next;
} e[maxn * 2];
void add(int from, int to) {
  cnt++;
  e[cnt] = {to, head[from]};
  head[from] = cnt;
}
void dfs(int u, int fa) {
  dp[u][0] = a[u];
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs(v, u);
  }
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    dp[u][0] += dp[v][k];
  }
  for (int dep = 1; dep <= k; dep++) {
    for (int i = head[u]; ~i; i = e[i].next) {
      int v1 = e[i].to;
      if (v1 == fa) continue;
      int sum = dp[v1][dep - 1];
      for (int j = head[u]; ~j; j = e[j].next) {
        int v2 = e[j].to;
        if (v2 == v1 || v2 == fa) continue;
        sum += dp[v2][max((k - 1) - (dep - 1), dep - 1)];
      }
      dp[u][dep] = max(dp[u][dep], sum);
    }
  }
  for (int dep = k - 1; dep >= 0; dep--)
    dp[u][dep] = max(dp[u][dep], dp[u][dep + 1]);
}
int main(void) {
  memset(head, -1, sizeof(head));
  cnt = 0;
  memset(dp, 0, sizeof(dp));
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  printf("%d\n", dp[1][0]);
}