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#include <bits/stdc++.h>
using namespace std;
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#define int long long
#define endl "\n"
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const int N = 200010, M = N << 1, K = 21;
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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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}
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int n, q;
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int dp[N][K];
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void dfs1(int u, int fa) {
for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
if (v == fa) continue;
dfs1(v, u);
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for (int j = 1; j <= q; j++) dp[u][j] += dp[v][j - 1]; // 第一遍dp
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}
}
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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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// 在第一次遍历时 dp[1][2] 包括了 dp[2][1] 2的子树权值
// 然鹅 ans在统计dp[2][3] 的时候也加上了 dp[2][1] 2的子树权值
// 第二次遍历 dp[2][3] 又加上了 dp[2][1];
// 所以需要简单容斥一下;
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for (int j = q; j >= 2; j--)
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dp[v][j] -= dp[v][j - 2]; // 简单容斥
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for (int j = 1; j <= q; j++)
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dp[v][j] += dp[u][j - 1]; // 第二遍dp
dfs2(v, u);
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}
}
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signed main() {
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// 初始化链式前向星
memset(h, -1, sizeof h);
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cin >> n >> q;
for (int i = 1; i < n; i++) {
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int a, b;
cin >> a >> b;
add(a, b), add(b, a);
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}
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for (int i = 1; i <= n; i++) cin >> dp[i][0]; // 每个节点往外0距离就是它本身的权值
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dfs1(1, 0);
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dfs2(1, 0);
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for (int i = 1; i <= n; i++) {
int ans = 0;
for (int j = 0; j <= q; j++) ans += dp[i][j]; // ans统计答案
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cout << ans << endl;
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}
}