#include<bits/stdc++.h>
using namespace std;int n, m;const int A = 300000;struct Point {int s, t, len, lca;
} r[300005];struct Node {int v, nex;
} Edge[600005], Edge_st[600005], Edge_ed[600005], Edge_lca[600005];int h[300005], stot;
void add(int u, int v) {Edge[++stot] = (Node) {v, h[u]};h[u] = stot;
}int hst[300005];
void add_st(int u, int id) {Edge_st[++stot] = (Node) {id, hst[u]};hst[u] = stot;
}int hed[300005];
void add_ed(int u, int id) {Edge_ed[++stot] = (Node) {id, hed[u]};hed[u] = stot;
}int hlca[300005];
void add_lca(int u, int id) {Edge_lca[++stot] = (Node) {id, hlca[u]};hlca[u] = stot;
}int dep[300005], jum[300005][21];
void dfs(int u, int f) {dep[u] = dep[f] + 1;jum[u][0] = f;for(int p = 1; p <= 20; p ++)jum[u][p] = jum[jum[u][p - 1]][p - 1];for(int i = h[u]; i; i = Edge[i].nex) {int v = Edge[i].v;if(v == f)    continue;dfs(v, u);}
}int LCA(int u, int v) {if(dep[u] < dep[v])    swap(u, v);int t = dep[u] - dep[v];for(int p = 0; t; t >>= 1, p ++)if(t & 1)    u = jum[u][p];if(u == v)    return u;for(int p = 20; p >= 0; p --)if(jum[u][p] != jum[v][p])    u = jum[u][p], v = jum[v][p];return jum[u][0];
}int ans0[300005], ans[300005], w[300005], t1[600005], t2[600005];
void Dfs(int u, int f) {ans0[u] = t1[dep[u] + w[u]] + t2[w[u] - dep[u] + A];///之前的贡献 不算在这个点的范围 for(int i = h[u]; i; i = Edge[i].nex) {int v = Edge[i].v;if(v == f)    continue;Dfs(v, u);}//回溯过程中更新答案 for(int i = hst[u]; i; i = Edge_st[i].nex)    t1[dep[u]] ++;for(int i = hed[u]; i; i = Edge_ed[i].nex)    t2[r[Edge_ed[i].v].len - dep[u] + A] ++;ans[u] = t1[dep[u] + w[u]] + t2[w[u] - dep[u] + A];////新增后的贡献（包括原来的） for(int i = hlca[u]; i; i = Edge_lca[i].nex) {int id = Edge_lca[i].v;t1[dep[r[id].s]] --;t2[r[id].len - dep[r[id].t] + A] --;/////以u为lca的所有路径的贡献在回溯回去时都没用了
    }
}int main() {scanf("%d%d", &n, &m);for(int i = 1; i < n; i ++) {int u, v;scanf("%d%d", &u, &v);add(u, v);    add(v, u);}dfs(1, 0);for(int i = 1; i <= n; i ++)    scanf("%d", &w[i]);for(int i = 1; i <= m; i ++) {int s, t;scanf("%d%d", &s, &t);r[i].s = s, r[i].t = t;r[i].lca = LCA(s, t);r[i].len = dep[s] + dep[t] - dep[r[i].lca] * 2;}stot = 0;for(int i = 1; i <= m; i ++)    add_st(r[i].s, i);stot = 0;for(int i = 1; i <= m; i ++)    add_ed(r[i].t, i);stot = 0;for(int i = 1; i <= m; i ++)    add_lca(r[i].lca, i);Dfs(1, 0);for(int i = 1; i <= m; i ++)if(dep[r[i].lca] + w[r[i].lca] == dep[r[i].s])    ans[r[i].lca] --;////如果lca可以观察到自己这条路，那它在两段路中多算了一次贡献 for(int i = 1; i <= n; i ++)    printf("%d ", ans[i] - ans0[i]);return 0;
}