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#include <bits/stdc++.h>
using namespace std;
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const int N = 1e5 + 10, M = N << 1;
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const int K = 25;
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// 链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
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int f[N][K]; // f[i][j]:如果根是1号节点时i号节点最远走j步可以获取到的所有点权和
int g[N][K];
int val[N]; // 点权数组
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int n, k;
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void dfs1(int u, int fa) {
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// 初始化当遍历到u节点时u的拆分状态中最起码包含了自己的点权值
for (int i = 0; i <= k; i++) f[u][0] = val[u];
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// 枚举u的每一个子节点
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for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
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if (v == fa) continue; // 如果是u的父亲那么就跳过,保证只访问u的孩子
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// 递归,利用子更新父
dfs1(v, u); // 递归填充v节点的信息
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// 在填充完子节点v的统计信息后利用儿子们的填充信息完成父亲节点信息的填充
// for(j=1,j<k,j++): 填充f[u]的每一个子状态孩子们的j=1层汇集的数据累加在一起就是f[u][j]的数据
// 最多计算k层足够用了
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for (int j = 1; j <= k; j++) f[u][j] += f[v][j - 1];
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}
}
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// 换根dp
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void dfs2(int u, int fa) {
for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
if (v == fa) continue;
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g[v][0] = val[v]; // 走0步,只有自己一个点
g[v][1] = f[v][1] + val[u]; // 走1步,包含自己下面子树一层+父节点
// 如果走2步及以上最多k步以内
for (int j = 2; j <= k; j++) g[v][j] = f[v][j] + g[u][j - 1] - f[v][j - 2];
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// 再递归,利用父更新子
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dfs2(v, u);
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}
}
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int main() {
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// 初始化链式前向星
memset(h, -1, sizeof h);
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cin >> n >> k;
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for (int i = 1; i < n; i++) { // n-1条边
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int a, b;
cin >> a >> b;
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add(a, b), add(b, a);
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}
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for (int i = 1; i <= n; i++) cin >> val[i]; // 点权
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// 1、自底向上
dfs1(1, 0);
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// 2、换根dp
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for (int i = 0; i <= k; i++) g[1][i] = f[1][i];
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dfs2(1, 0);
// 输出结果
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for (int i = 1; i <= n; i++) cout << g[i][k] << endl;
return 0;
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}