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 = 2e5 + 10;
long long n, st[MAXN], fi[MAXN], q, tim, answer[MAXN], lazy[4 * MAXN];
vector<int> adj[MAXN], lis[MAXN];
pair<int, int> qu[MAXN], seg[4 * MAXN];
void dfs_(int v, int par) {
  st[v] = ++tim;
  for (auto i : adj[v]) {
    if (i != par) {
      dfs_(i, v);
    }
  }
  fi[v] = tim;
  return;
}
void built(int l = 1, int r = n + 1, int id = 1) {
  if (l == (r - 1)) {
    seg[id] = {0, 1};
    return;
  }
  int mid = l + r >> 1;
  built(l, mid, id << 1);
  built(mid, r, id << 1 | 1);
  seg[id] = {0, r - l};
  return;
}
void seg_upt(int b, int e, int val, int l = 1, int r = n + 1, int id = 1) {
  if (l >= e || r <= b) {
    return;
  }
  if (l >= b && r <= e) {
    lazy[id] += val;
    return;
  }
  int mid = l + r >> 1;
  seg_upt(b, e, val, l, mid, id << 1);
  seg_upt(b, e, val, mid, r, id << 1 | 1);
  if ((seg[id << 1].first + lazy[id << 1]) ==
      (seg[id << 1 | 1].first + lazy[id << 1 | 1])) {
    seg[id] = {seg[id << 1].first + lazy[id << 1],
               seg[id << 1].second + seg[id << 1 | 1].second};
  } else {
    seg[id] = min(
        pair<int, int>(seg[id << 1].first + lazy[id << 1], seg[id << 1].second),
        pair<int, int>(seg[id << 1 | 1].first + lazy[id << 1 | 1],
                       seg[id << 1 | 1].second));
  }
  return;
}
pair<int, int> seg_get() {
  return pair<int, int>(seg[1].first + lazy[1], seg[1].second);
}
void dfs(int v, int par) {
  for (auto i : lis[v]) {
    int a = qu[i].first, b = qu[i].second;
    seg_upt(st[a], fi[a] + 1, 1);
    seg_upt(st[b], fi[b] + 1, 1);
  }
  pair<int, int> ans = seg_get();
  answer[v] = n - 1 - (ans.first ? 0 : ans.second);
  for (auto i : adj[v]) {
    if (i != par) {
      dfs(i, v);
    }
  }
  for (auto i : lis[v]) {
    int a = qu[i].first, b = qu[i].second;
    seg_upt(st[a], fi[a] + 1, -1);
    seg_upt(st[b], fi[b] + 1, -1);
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs_(1, -1);
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    qu[i] = {u, v};
    lis[u].push_back(i);
    lis[v].push_back(i);
  }
  built();
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    if (answer[i] == -1) {
      answer[i]++;
    }
    cout << answer[i] << ' ';
  }
  return 0;
}
