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 = 3e5 + 10;
int pre[maxn];
long long sum[maxn], t_ans[maxn];
struct node {
  int u;
  int v;
  long long w;
} edge[maxn];
struct qqq {
  long long val;
  int pos;
} query[maxn];
int find(int x) { return pre[x] == x ? x : pre[x] = find(pre[x]); }
bool cmp(node x, node y) { return x.w < y.w; }
bool cmp1(qqq x, qqq y) { return x.val < y.val; }
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d%I64d", &edge[i].u, &edge[i].v, &edge[i].w);
  }
  sort(edge + 1, edge + n, cmp);
  for (int i = 1; i <= q; i++) {
    scanf("%I64d", &query[i].val);
    query[i].pos = i;
  }
  sort(query + 1, query + 1 + q, cmp1);
  for (int i = 1; i <= n; i++) {
    pre[i] = i;
    sum[i] = 1;
  }
  int len = 1;
  long long temp = 0;
  for (int i = 1; i <= q; i++) {
    while (query[i].val >= edge[len].w && len <= n - 1) {
      int x = find(edge[len].v), y = find(edge[len].u);
      if (x != y) {
        temp -= (sum[x] * (sum[x] - 1) / 2);
        temp -= (sum[y] * (sum[y] - 1) / 2);
        sum[y] += sum[x];
        pre[x] = y;
        temp += (sum[y] * (sum[y] - 1) / 2);
      }
      len++;
    }
    t_ans[query[i].pos] = temp;
  }
  for (int i = 1; i <= q; i++) {
    cout << t_ans[i] << " ";
  }
  return 0;
}