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#include <bits/stdc++.h>
using namespace std;
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const int N = 500010, M = N << 1;
#define int long long
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int n, m, k;
#define INF 1e18
int sz[N]; // 以u为根节点是否有人
int g[N]; // 从u出发把u子树上的人都送回家再回到u所需要的时间
vector<pair<int, int>> vec[N];
int max1[N]; // u的子树中最长链的长度
int max2[N]; // u的子树中次长链
int up[N]; // 不在u的子树内距离u最远的那个人的家到u的距离
int id[N]; // 最长链条是哪条链
int ans[N]; // 从u出发把所有点都送回家再回到u的结果
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// ans[i] -max(up[i],max1[i]);
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void dfs1(int u, int fa) {
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for (auto ts : vec[u]) {
int v = ts.first;
int w = ts.second;
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if (v == fa) continue;
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dfs1(v, u);
if (sz[v]) {
g[u] += g[v] + 2 * w;
int now = max1[v] + w;
if (now >= max1[u]) {
max2[u] = max1[u];
max1[u] = now;
id[u] = v;
} else if (now >= max2[u]) {
max2[u] = now;
}
}
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sz[u] += sz[v];
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}
}
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void dfs2(int u, int fa) {
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for (auto ts : vec[u]) {
int v = ts.first;
int w = ts.second;
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if (v == fa) continue;
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if (sz[v] == k) {
ans[v] = g[v];
up[v] = 0;
} else if (sz[v] == 0) {
ans[v] = ans[u] + 2 * w;
up[v] = max(up[u], max1[u]) + w;
} else if (sz[v] && sz[v] != k) {
ans[v] = ans[u];
if (id[u] == v) {
up[v] = max(max2[u], up[u]) + w;
} else
up[v] = max(up[u], max1[u]) + w;
}
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dfs2(v, u);
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}
}
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signed main() {
cin >> n >> k;
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for (int i = 1; i < n; i++) {
int a, b, c;
cin >> a >> b >> c;
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vec[a].push_back({b, c});
vec[b].push_back({a, c});
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}
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for (int i = 1; i <= k; i++) {
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int x;
cin >> x;
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sz[x] = 1;
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}
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dfs1(1, 0);
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ans[1] = g[1];
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dfs2(1, 0);
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int minn = INF;
for (int u = 1; u <= n; u++) cout << ans[u] - max(up[u], max1[u]) << endl;
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}