1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define R register int
  4 #define AC 501000
  5 #define ac 910000//要开这么大。。。
  6 #define getchar() *o++
  7 char READ[12000100], *o = READ;
  8 int n, m;
  9 int w[AC], ans[AC], s[AC], t[AC], dis[AC], dep[AC], may[ac], id[ac];
 10 int Head[AC], Next[ac], date[ac], tot;
 11 struct edge{
 12     int Head[AC], Next[ac], date[ac], tot;
 13     inline void add(int f, int w)
 14     {
 15         date[++tot] = w, Next[tot] = Head[f], Head[f] = tot;
 16         date[++tot] = f, Next[tot] = Head[w], Head[w] = tot;
 17     }
 18 }E1,E2,E3;//询问的边(LCA)，查询的边(ans)
 19
 20 inline int read()
 21 {
 22     int x = 0; char c = getchar();
 23     while(c > '9' || c < '0') c = getchar();
 24     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
 25     return x;
 26 }
 27
 28 inline void add(int f, int w)
 29 {
 30     date[++tot] = w, Next[tot] = Head[f], Head[f] = tot;
 31     date[++tot] = f, Next[tot] = Head[w], Head[w] = tot;
 32 }
 33
 34 inline void add1(int f, int w, int S)
 35 {
 36     E3.date[++tot] = w, E3.Next[tot] = E3.Head[f], E3.Head[f] = tot, may[tot] = S;
 37 }
 38
 39 inline void add2(int f, int w)//这里只能连单向边
 40 {
 41     E2.date[++tot] = w, E2.Next[tot] = E2.Head[f], E2.Head[f] = tot;
 42 }
 43
 44 void pre()
 45 {
 46     int a, b;
 47     n = read(), m = read();
 48     for(R i = 1; i < n; i++)
 49     {
 50         a = read(), b = read();
 51         add(a, b);
 52     }
 53     for(R i = 1; i <= n; i++) w[i] = read();
 54     E1.tot = E2.tot = tot = 1;
 55     for(R i = 1; i <= m; i++)
 56     {
 57         s[i] = read(), t[i] = read();
 58         E1.add(s[i], t[i]);
 59         add1(s[i], i, 0);//标记为开始节点
 60         add1(t[i], i, 1);//标记为结束节点
 61         id[E1.tot - 1] = id[E1.tot] = i;
 62     }
 63 }
 64
 65 struct get_LCA{
 66     int father[AC], LCA[ac]; bool z[AC];
 67
 68     inline int find(int x)
 69     {
 70          return (x == father[x]) ? x : father[x] = find(father[x]);
 71     }
 72
 73     void dfs(int x)
 74     {
 75         int now;
 76         z[x] = true;
 77         for(R i = Head[x]; i; i = Next[i])
 78         {
 79             now = date[i];
 80             if(z[now]) continue;
 81             dep[now] = dep[x] + 1;
 82             dfs(now);
 83             father[now] = x;//访问完就要改父亲了
 84         }
 85         for(R i = E1.Head[x]; i; i = E1.Next[i])
 86         {
 87             now = E1.date[i];
 88             if(z[now] && !LCA[i ^ 1])
 89             {
 90             //    printf("%d %d\n", x, now);
 91                 LCA[i] = find(now);
 92                 dis[id[i]] = dep[x] - dep[LCA[i]] + dep[now] - dep[LCA[i]];
 93                 add2(LCA[i], id[i]);//将LCA和询问联系起来
 94             }
 95         }
 96     }
 97
 98     void getLCA()
 99     {
100         for(R i = 1; i <= n; i++) father[i] = i;
101         dep[1] = 1;
102         dfs(1);
103     }
104 }LCA;
105
106 #define k 600000
107 struct difference{
108     int bu[ac], b[ac * 3];
109     void dfs(int x, int fa)
110     {
111         int tmp = bu[dep[x] + w[x]], rnt = b[w[x] - dep[x] + k], now;
112         for(R i = E3.Head[x]; i; i = E3.Next[i])//加上当前节点的贡献
113         {
114             now = E3.date[i];
115             if(!may[i]) ++bu[dep[x]];//如果是开始节点
116             else ++b[dis[now] - dep[x] + k];//等式两边同时+k
117         }
118         for(R i = Head[x]; i; i = Next[i])//遍历子树
119         {
120             now = date[i];
121             if(now == fa) continue;
122             dfs(now, x);
123         }
124         ans[x] = bu[dep[x] + w[x]] - tmp + b[w[x] - dep[x] + k] - rnt;
125         for(R i = E2.Head[x]; i; i = E2.Next[i])//查看当前节点是哪些点对的LCA
126         {
127             now = E2.date[i];
128             if(dep[s[now]] == dep[x] + w[x]) --ans[x];//如果造成了贡献，那么必定是双倍，因此要减去
129             --bu[dep[s[now]]];//减去贡献
130             --b[dis[now] - dep[t[now]] + k];//等式两边同时+k!
131         }
132     }
133
134 }get;
135
136 void work()
137 {
138     for(R i = 1; i <= n; i++) printf("%d ", ans[i]);
139     printf("\n");
140 }
141
142 int main()
143 {
144     freopen("in.in", "r", stdin);
145     fread(READ, 1, 12000000, stdin);
146     pre();
147     LCA.getLCA();
148     get.dfs(1, 0);
149     work();
150     fclose(stdin);
151     return 0;
152 }