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 readint() {
  int a = 0;
  char c = getchar(), f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -f;
  for (; '0' <= c && c <= '9'; c = getchar())
    a = (a << 3) + (a << 1) + (c ^ 48);
  return a * f;
}
template <class T>
void getMax(T& a, const T& b) {
  if (a < b) a = b;
}
template <class T>
void getMin(T& a, const T& b) {
  if (b < a) a = b;
}
const int MaxN = 205;
struct Edge {
  int to, nxt;
  Edge(int T = 0, int N = 0) { to = T, nxt = N; }
} edge[MaxN << 1];
int head[MaxN], cntEdge;
void addEdge(int a, int b) {
  edge[cntEdge] = Edge(b, head[a]);
  head[a] = cntEdge++;
  edge[cntEdge] = Edge(a, head[b]);
  head[b] = cntEdge++;
}
int n, k, tmp[MaxN], a[MaxN];
int dp[MaxN][MaxN];
void dfs(int x, int pre) {
  dp[x][0] = a[x];
  for (int i = head[x]; ~i; i = edge[i].nxt) {
    if (edge[i].to == pre) continue;
    int y = edge[i].to;
    dfs(y, x);
    for (int j = k; ~j; --j) tmp[j] = 0;
    for (int j = k; ~j; --j)
      for (int t = j + 1; t; --t)
        getMax(tmp[j], dp[y][t] + dp[x][min(j, k + 2 - t)]);
    getMax(tmp[k], dp[y][0] + dp[x][1]);
    for (int j = k; ~j; --j) dp[x][j] = tmp[j];
  }
}
int main() {
  n = readint(), k = readint();
  for (int i = 1; i <= n; ++i) {
    head[i] = -1;
    a[i] = readint();
  }
  for (int i = 1; i < n; ++i) addEdge(readint(), readint());
  dfs(1, -1);
  printf("%d\n", max(dp[1][0], dp[1][k]));
  return 0;
}