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 = 3e5 + 10, mod = 998244353;
const int INF = 1e8;
int n, m;
struct edge {
  int x, y, w;
} e[N];
struct node {
  int k, xh;
  long long ans;
} b[N];
bool cmp(edge p, edge q) { return p.w < q.w; }
bool mmp(node p, node q) { return p.k < q.k; }
bool mmmp(node p, node q) { return p.xh < q.xh; }
long long ans;
int a[N], s[N], ss[N];
int f(int x) { return x == a[x] ? x : a[x] = f(a[x]); }
void u(int x, int y) {
  x = f(x), y = f(y);
  ans += 1ll * s[x] * s[y];
  s[y] += s[x];
  a[x] = y;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) a[i] = i, s[i] = 1;
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].w);
  }
  sort(e + 1, e + n, cmp);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &b[i].k);
    b[i].xh = i;
  }
  sort(b + 1, b + 1 + m, mmp);
  int cnt = 0;
  for (int i = 1; i <= m; i++) {
    while (e[cnt + 1].w <= b[i].k && cnt < n - 1) {
      cnt++;
      u(e[cnt].x, e[cnt].y);
    }
    b[i].ans = ans;
  }
  sort(b + 1, b + 1 + m, mmmp);
  for (int i = 1; i <= m; i++) {
    printf("%lld ", b[i].ans);
  }
  return 0;
}