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 = (int)2e5 + 100;
const int INF = (int)1e9 + 100;
int n, m;
int pre[maxn], tot[maxn], Rank[maxn];
long long ans = 0;
struct node {
  int u, v, w;
  bool operator<(const node &b) const { return w < b.w; }
} g[maxn];
struct que {
  int ww, id;
  long long ans;
  bool operator<(const que &b) const { return ww < b.ww; }
} q[maxn];
int cmp(que a, que b) { return a.id < b.id; }
int find(int x) { return x == pre[x] ? x : pre[x] = find(pre[x]); }
void join(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  ans -= (1ll * tot[x] * (tot[x] - 1)) / 2;
  ans -= (1ll * tot[y] * (tot[y] - 1)) / 2;
  ans += (1ll * (tot[x] + tot[y]) * (tot[x] + tot[y] - 1)) / 2;
  if (Rank[x] > Rank[y]) {
    pre[y] = x;
    if (x != y) tot[x] += tot[y];
  } else {
    pre[x] = y;
    if (x != y) tot[y] += tot[x];
    if (Rank[x] == Rank[y]) Rank[y]++;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (n - 1); ++i)
    scanf("%d%d%d", &g[i].u, &g[i].v, &g[i].w);
  sort(g + 1, g + n);
  for (int i = (1); i <= (m); ++i) scanf("%d", &q[i].ww), q[i].id = i;
  sort(q + 1, q + 1 + m);
  for (int i = (1); i <= (n); ++i) pre[i] = i, tot[i] = 1;
  int pt = 1;
  for (int i = (1); i <= (m); ++i) {
    while (pt < n && g[pt].w <= q[i].ww) join(g[pt].u, g[pt].v), pt++;
    q[i].ans = ans;
  }
  sort(q + 1, q + 1 + m, cmp);
  for (int i = (1); i <= (m); ++i) printf("%lld ", q[i].ans);
}