The Little Elephant loves trees very much, he especially loves root trees.

He's got a tree consisting of n nodes (the nodes are numbered from 1 to n), with root at node number 1. Each node of the tree contains some list of numbers which initially is empty. 

The Little Elephant wants to apply m operations. On the i-th operation (1 ≤ i ≤ m) he first adds number i to lists of all nodes of a subtree with the root in node number ai, and then he adds number i to lists of all nodes of the subtree with root in node bi.

After applying all operations the Little Elephant wants to count for each node i number ci — the number of integers j (1 ≤ j ≤ n; j ≠ i), such that the lists of the i-th and the j-th nodes contain at least one common number.

Help the Little Elephant, count numbers ci for him.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of the tree nodes and the number of operations. 

Each of the following n - 1 lines contains two space-separated integers, ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), that mean that there is an edge between nodes number ui and vi. 

Each of the following m lines contains two space-separated integers, ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi), that stand for the indexes of the nodes in the i-th operation.

It is guaranteed that the given graph is an undirected tree.

Output

In a single line print n space-separated integers — c1, c2, ..., cn.

Examples

Input

5 1
1 2
1 3
3 5
3 4
2 3


Output

0 3 3 3 3 

Input

11 3
1 2
2 3
2 4
1 5
5 6
5 7
5 8
6 9
8 10
8 11
2 9
3 6
2 8


Output

0 6 7 6 0 2 0 5 4 5 5 
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
struct edge {
  int v;
  int next;
} edge[maxn << 1];
int head[maxn], tot, num;
int s[maxn], e[maxn];
vector<int> que[maxn];
int n, m, ans[maxn];
void addedge(int u, int v) {
  edge[tot].v = v;
  edge[tot].next = head[u];
  head[u] = tot++;
  edge[tot].v = u;
  edge[tot].next = head[v];
  head[v] = tot++;
}
struct node {
  int l, r;
  int length() { return r - l + 1; }
};
struct segtree {
  node seg[maxn << 2];
  int cnt[maxn << 2];
  int lazy[maxn << 2];
  void pushup(int o) {
    if (lazy[o])
      cnt[o] = seg[o].length();
    else {
      if (seg[o].l == seg[o].r)
        cnt[o] = 0;
      else
        cnt[o] = cnt[o << 1] + cnt[o << 1 | 1];
    }
  }
  void build(int o, int l, int r) {
    seg[o].l = l, seg[o].r = r, cnt[o] = lazy[o] = 0;
    if (l == r) return;
    int m = (l + r) >> 1;
    build(o << 1, l, m);
    build(o << 1 | 1, m + 1, r);
  }
  void update(int o, int l, int r, int v) {
    if (l <= seg[o].l && seg[o].r <= r) {
      lazy[o] += v;
      pushup(o);
      return;
    }
    int m = (seg[o].l + seg[o].r) >> 1;
    if (l <= m) update(o << 1, l, r, v);
    if (r > m) update(o << 1 | 1, l, r, v);
    pushup(o);
  }
} tree;
void solve(int u, int pre) {
  for (int i = 0; i < que[u].size(); i++) {
    int v = que[u][i];
    tree.update(1, s[v], e[v], 1);
  }
  ans[u] = tree.cnt[1];
  if (ans[u]) ans[u]--;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (v == pre) continue;
    solve(v, u);
  }
  for (int i = 0; i < que[u].size(); i++) {
    int v = que[u][i];
    tree.update(1, s[v], e[v], -1);
  }
}
void dfs(int u, int pre) {
  s[u] = ++num;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (v == pre) continue;
    dfs(v, u);
  }
  e[u] = num;
}
void init() {
  tot = 0;
  num = 0;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; i++) que[i].clear();
}
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    init();
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      addedge(u, v);
    }
    for (int i = 0; i < m; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      que[a].push_back(b), que[a].push_back(a);
      que[b].push_back(a), que[b].push_back(b);
    }
    dfs(1, -1);
    tree.build(1, 1, n);
    solve(1, -1);
    for (int i = 1; i < n; i++) printf("%d ", ans[i]);
    printf("%d\n", ans[n]);
  }
}
