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 maxn = 2e5 + 5;
struct node {
  int u, v, w;
  friend bool operator<(const node &x, const node &y) { return x.w < y.w; }
};
struct query {
  int q, id;
  long long ans;
};
vector<node> edges;
vector<query> querys;
int pa[maxn], sz[maxn];
int find(int u) {
  if (pa[u] == u) return u;
  return pa[u] = find(pa[u]);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    pa[i] = i;
    sz[i] = 1;
  }
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    edges.push_back({u, v, w});
  }
  sort(edges.begin(), edges.end());
  for (int i = 0; i < m; i++) {
    int q;
    scanf("%d", &q);
    querys.push_back({q, i, 0});
  }
  sort(querys.begin(), querys.end(),
       [](const query &x, const query &y) { return x.q < y.q; });
  long long res = 0;
  int id = 0;
  for (int i = 0; i < m; i++) {
    while (id < n - 1 && edges[id].w <= querys[i].q) {
      int fu = find(edges[id].u);
      int fv = find(edges[id].v);
      if (fu != fv) {
        pa[fu] = fv;
        res = (res - 1LL * sz[fu] * (sz[fu] - 1) / 2 -
               1LL * sz[fv] * (sz[fv] - 1) / 2 +
               1LL * (sz[fu] + sz[fv]) * (sz[fu] + sz[fv] - 1) / 2);
        sz[fv] += sz[fu];
      }
      id++;
    }
    querys[i].ans = res;
  }
  sort(querys.begin(), querys.end(),
       [](const query &x, const query &y) { return x.id < y.id; });
  for (int i = 0; i < m; i++) printf("%lld ", querys[i].ans);
  return 0;
}