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;
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long readll() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int MAXN = 2e5 + 5;
struct edge {
  int u, v, w;
} e[MAXN];
struct query {
  int x, id;
} q[MAXN];
bool cmpe(edge a, edge b) { return a.w < b.w; }
bool cmpq(query a, query b) { return a.x < b.x; }
int n, m, fa[MAXN];
long long sz[MAXN], t[MAXN], ans[MAXN];
int findFather(int x) {
  if (x == fa[x]) return x;
  return fa[x] = findFather(fa[x]);
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i < n; ++i) e[i].u = read(), e[i].v = read(), e[i].w = read();
  for (int i = 1; i <= m; ++i) q[i].x = read(), q[i].id = i;
  sort(e + 1, e + n, cmpe);
  sort(q + 1, q + m + 1, cmpq);
  for (int i = 1; i <= n; ++i) fa[i] = i, sz[i] = 1;
  int j = 1;
  long long res = 0;
  for (int i = 1; i <= m; ++i) {
    while (j < n && e[j].w <= q[i].x) {
      int u = findFather(e[j].u), v = findFather(e[j].v);
      if (u != v) {
        if (sz[u] > sz[v]) swap(u, v);
        res -= t[u] + t[v];
        fa[u] = v;
        sz[v] += sz[u];
        t[v] = sz[v] * (sz[v] - 1) / 2;
        res += t[v];
      }
      ++j;
    }
    ans[q[i].id] = res;
  }
  for (int i = 1; i <= m; ++i) printf("%lld ", ans[i]);
  puts("");
  return 0;
}