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;
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const long long MAXN = 1e5 + 10;
long long q, n, u, v, a, b, t, st[MAXN], fn[MAXN], C[MAXN], seg[2][MAXN << 2],
    lazy[MAXN << 2];
vector<long long> adj[MAXN], query[MAXN];
void dfs(long long u, long long p = -1) {
  st[u] = t++;
  for (auto v : adj[u])
    if (v != p) dfs(v, u);
  fn[u] = t;
  return;
}
void build(long long id = 1, long long second = 0, long long e = n) {
  if (e - second < 2) {
    seg[1][id] = 1;
    return;
  }
  long long mid = (second + e) >> 1;
  build(id << 1, second, mid), build(id << 1 | 1, mid, e),
      seg[1][id] = e - second;
  return;
}
void shift(long long id) {
  if (!lazy[id]) return;
  seg[0][id << 1] += lazy[id], seg[0][id << 1 | 1] += lazy[id],
      lazy[id << 1] += lazy[id], lazy[id << 1 | 1] += lazy[id];
  lazy[id] = 0;
  return;
}
void add(long long l, long long r, long long x, long long id = 1,
         long long second = 0, long long e = n) {
  if (e <= l or r <= second) return;
  if (l <= second and e <= r) {
    seg[0][id] += x, lazy[id] += x;
    return;
  }
  shift(id);
  long long mid = (second + e) >> 1;
  if (l < mid) add(l, r, x, id << 1, second, mid);
  if (mid < r) add(l, r, x, id << 1 | 1, mid, e);
  seg[0][id] = min(seg[0][id << 1], seg[0][id << 1 | 1]);
  if (seg[0][id << 1] == seg[0][id << 1 | 1])
    seg[1][id] = seg[1][id << 1] + seg[1][id << 1 | 1];
  else {
    if (seg[0][id << 1] < seg[0][id << 1 | 1])
      seg[1][id] = seg[1][id << 1];
    else
      seg[1][id] = seg[1][id << 1 | 1];
  }
  return;
}
void DFS(long long u, long long p = -1) {
  for (auto v : query[u]) add(st[v], fn[v], 1), add(st[u], fn[u], 1);
  if (seg[0][1])
    C[u] = n - 1;
  else
    C[u] = n - seg[1][1] - 1;
  for (auto v : adj[u])
    if (v != p) DFS(v, u);
  for (auto v : query[u]) add(st[v], fn[v], -1), add(st[u], fn[u], -1);
  return;
}
void input() {
  cin >> n >> q;
  for (long long i = 0; i < n - 1; i++)
    cin >> u >> v, u--, v--, adj[u].push_back(v), adj[v].push_back(u);
  dfs(0);
  while (q--)
    cin >> a >> b, a--, b--, query[a].push_back(b), query[b].push_back(a);
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input(), build(), DFS(0);
  for (long long i = 0; i < n; i++)
    if (C[i] < 0) C[i] = 0;
  for (long long i = 0; i < n; i++) cout << C[i] << " ";
  return cout << '\n', 0;
}
