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 x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
struct cow {
  int v, next;
} E[10009 << 1];
int head[10009];
int n, k;
int tot = 0;
void add(int u, int v) {
  E[++tot].v = v;
  E[tot].next = head[u];
  head[u] = tot;
  E[++tot].v = u;
  E[tot].next = head[v];
  head[v] = tot;
}
int q[10009];
bool vis[10009];
void bfs() {
  int h = 1, t = 1;
  q[1] = vis[1] = 1;
  while (h <= t) {
    int u = q[h++];
    for (int i = head[u]; i; i = E[i].next) {
      int v = E[i].v;
      if (!vis[v]) {
        q[++t] = v;
        vis[v] = true;
      }
    }
  }
}
int A[10009], val;
void dfs(int u, int fa, int dis) {
  if (dis > k) return;
  A[u] -= val;
  for (int i = head[u]; i; i = E[i].next) {
    int v = E[i].v;
    if (v == fa) continue;
    dfs(v, u, dis + 1);
  }
}
int main() {
  int u, v;
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) A[i] = read();
  for (int i = 1; i < n; i++) {
    u = read();
    v = read();
    add(u, v);
  }
  int ans = 0;
  bfs();
  for (int i = n; i >= 1; i--) {
    u = q[i];
    val = A[u];
    if (val <= 0) continue;
    ans += val;
    dfs(u, 0, 0);
  }
  printf("%d\n", ans);
  return 0;
}