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;
int n, k, a[205], tot, head[205], q[205], num;
bool vis[205];
struct node {
  int v, next;
} e[205 << 1];
void dfs(int x, int pre, int dp) {
  if (dp > k) return;
  a[x] -= num;
  for (int i = head[x]; i != -1; i = e[i].next) {
    int v = e[i].v;
    if (v == pre) continue;
    dfs(v, x, dp + 1);
  }
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d %d", &x, &y);
    e[tot] = node{y, head[x]}, head[x] = tot++;
    e[tot] = node{x, head[y]}, head[y] = tot++;
  }
  int l = 0, r = 1;
  q[++l] = 1, vis[1] = true;
  while (l <= r) {
    int x = q[l++];
    for (int i = head[x]; i != -1; i = e[i].next) {
      int v = e[i].v;
      if (!vis[v]) q[++r] = v, vis[v] = true;
    }
  }
  int ans = 0;
  for (int i = n; i >= 1; i--) {
    if (a[q[i]] <= 0) continue;
    num = a[q[i]];
    ans += num;
    dfs(q[i], 0, 0);
  }
  printf("%d\n", ans);
  return 0;
}