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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int ret = 0, f = 0;
  char c = gc();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = gc();
  }
  while (isdigit(c)) {
    ret = ret * 10 + c - 48;
    c = gc();
  }
  if (f) return -ret;
  return ret;
}
const int maxn = 205;
int val[maxn];
vector<int> g[maxn];
int dp[maxn][maxn];
int n, k;
void dfs(int u, int fa) {
  dp[u][0] = val[u];
  for (auto v : g[u]) {
    if (v == fa) continue;
    dfs(v, u);
  }
  for (auto v : g[u]) {
    if (v == fa) continue;
    dp[u][0] += dp[v][k];
  }
  for (int i = 1; i < n; i++) {
    for (auto v1 : g[u]) {
      if (v1 == fa) continue;
      int tmp = dp[v1][i - 1];
      for (auto v2 : g[u]) {
        if (v2 == v1 || v2 == fa) continue;
        tmp += dp[v2][max(i - 1, k - i)];
      }
      dp[u][i] = max(dp[u][i], tmp);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    dp[u][i] = max(dp[u][i], dp[u][i + 1]);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &val[i]);
  }
  for (int i = (1); i <= (n - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 1);
  printf("%d\n", dp[1][0]);
  return 0;
}