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 long long MAXN = 2 * 1e5 + 100;
long long n, m, ans[MAXN], now, fa[MAXN];
long long si[MAXN], cnt;
struct node {
  long long u, v, w;
} sh[MAXN];
struct query {
  long long num, wh;
} q[MAXN];
bool cmp(node a, node b) { return a.w < b.w; }
bool cmp1(query a, query b) { return a.num < b.num; }
long long find(long long x) {
  if (fa[x] == x) return fa[x];
  fa[x] = find(fa[x]);
  return fa[x];
}
int main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i < n; i++)
    scanf("%lld%lld%lld", &sh[i].u, &sh[i].v, &sh[i].w);
  for (long long i = 1; i <= m; i++) scanf("%lld", &q[i].num);
  for (long long i = 1; i <= m; i++) q[i].wh = i;
  sort(q + 1, q + 1 + m, cmp1);
  sort(sh + 1, sh + n, cmp);
  for (long long i = 1; i <= n; i++) fa[i] = i, si[i] = 1;
  now = 1;
  cnt = 0;
  for (long long i = 1; i <= m; i++) {
    while (sh[now].w <= q[i].num && now <= n) {
      long long fu, fv;
      fu = find(sh[now].u);
      fv = find(sh[now].v);
      if (fu != fv) {
        cnt -= (si[fu] * (si[fu] - 1)) / 2 + (si[fv] * (si[fv] - 1)) / 2;
        fa[fu] = fv;
        si[fv] += si[fu];
        cnt += (si[fv] * (si[fv] - 1)) / 2;
      }
      now++;
    }
    ans[q[i].wh] = cnt;
  }
  for (long long i = 1; i <= m; i++) printf("%lld ", ans[i]);
  printf("\n");
}