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 N = 2e5 + 50;
struct Edge {
  int u, v, w;
  bool operator<(const Edge &b) const { return w < b.w; }
} edge[N];
struct node {
  int id, w;
  bool operator<(const node &b) const { return w < b.w; }
} q[N];
long long ans[N];
int fa[N], num[N];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
  }
  for (int i = 1; i <= m; i++) scanf("%d", &q[i].w), q[i].id = i;
  sort(q + 1, q + m + 1);
  sort(edge + 1, edge + n);
  for (int i = 1; i <= n; i++) {
    fa[i] = i;
    num[i] = 1;
  }
  int now = 1;
  long long res = 0;
  for (int i = 1; i <= m; i++) {
    while (edge[now].w <= q[i].w && now < n) {
      int x = find(edge[now].u);
      int y = find(edge[now].v);
      res += 1ll * num[x] * num[y];
      fa[x] = y;
      num[y] += num[x];
      now++;
    }
    ans[q[i].id] = res;
  }
  for (int i = 1; i <= m; i++) {
    printf("%lld ", ans[i]);
  }
  return 0;
}