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;
long long a[405];
long long n;
long long kk;
struct edge {
  long long des;
  long long next;
} e[405 << 1];
long long cnt;
long long head[405];
long long dp[405][405];
long long tmp[405];
long long length[405];
void ins(long long u, long long v) {
  ++cnt;
  e[cnt].des = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long f) {
  dp[u][0] = a[u];
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].des;
    if (v == f) continue;
    dfs(v, u);
    long long newlength = max(length[u], length[v] + 1);
    for (long long j = 0; j <= newlength; ++j) tmp[j] = -1;
    for (long long j = 0; j <= length[u]; ++j) {
      for (long long k = kk + 1 - j - 1; k <= length[v]; ++k) {
        tmp[min(j, k + 1)] = max(tmp[min(j, k + 1)], dp[u][j] + dp[v][k]);
      }
    }
    for (long long j = 0; j <= length[v]; ++j)
      tmp[j + 1] = max(tmp[j + 1], dp[v][j]);
    length[u] = newlength;
    for (long long j = 0; j <= newlength; ++j) dp[u][j] = max(tmp[j], dp[u][j]);
    for (long long j = newlength - 1; j >= 0; --j)
      dp[u][j] = max(dp[u][j], dp[u][j + 1]);
  }
}
int main() {
  scanf("%lld %lld", &n, &kk);
  for (long long i = 1; i <= n; ++i) scanf("%lld", a + i);
  for (long long i = 1; i < n; ++i) {
    long long u;
    long long v;
    scanf("%lld %lld", &u, &v);
    ins(u, v);
    ins(v, u);
  }
  dfs(1, 0);
  printf("%lld\n", dp[1][0]);
  return 0;
}