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 = 100000 + 10;
long long seg[6 * maxn];
vector<int> adj[maxn];
vector<int> adj1[maxn];
int st[maxn];
int en[maxn];
bool mark[maxn];
int cnt[6 * maxn];
long long ans[maxn];
vector<int> path;
int n, m;
void Add(int s, int e, int tag, int l, int r, int val) {
  if (s >= l && e <= r) {
    cnt[tag] += val;
    if (cnt[tag] > 0) {
      seg[tag] = e - s + 1;
      return;
    } else {
      seg[tag] = seg[tag * 2 + 1] + seg[2 * tag];
      return;
    }
  }
  if (s > r || e < l) return;
  int m = (s + e) / 2;
  Add(s, m, tag * 2, l, r, val);
  Add(m + 1, e, tag * 2 + 1, l, r, val);
  if (cnt[tag])
    seg[tag] = e - s + 1;
  else
    seg[tag] = seg[tag * 2] + seg[tag * 2 + 1];
  return;
}
void Dfs1(int v, int par) {
  path.push_back(v);
  st[v] = path.size();
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != par) Dfs1(u, v);
  }
  en[v] = path.size();
  return;
}
void Dfs2(int v, int par) {
  for (int i = 0; i < adj1[v].size(); i++) {
    int l = st[adj1[v][i]];
    int r = en[adj1[v][i]];
    Add(1, n, 1, l, r, 1);
  }
  if (mark[v]) {
    int l = st[v];
    int r = en[v];
    Add(1, n, 1, l, r, 1);
  }
  ans[v] = seg[1] - 1;
  if (ans[v] < 0) ans[v] = 0;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != par) Dfs2(u, v);
  }
  for (int i = 0; i < adj1[v].size(); i++) {
    int l = st[adj1[v][i]];
    int r = en[adj1[v][i]];
    Add(1, n, 1, l, r, -1);
  }
  if (mark[v]) {
    int l = st[v];
    int r = en[v];
    Add(1, n, 1, l, r, -1);
  }
  return;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  Dfs1(1, 0);
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    adj1[a].push_back(b);
    adj1[b].push_back(a);
    mark[a] = true;
    mark[b] = true;
  }
  Dfs2(1, 0);
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
