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 = 1e5 + 10, maxsg = (1 << 18) + 10;
int n, m, t, st[maxn], ft[maxn], seg[maxsg], lazy[maxsg], low, high, ans[maxn];
vector<int> adj[maxn], q[maxn];
void dfs(int v = 0, int par = -1) {
  st[v] = t++;
  for (auto u : adj[v])
    if (u != par) dfs(u, v);
  ft[v] = t;
}
void update(int val, int id = 1, int l = 0, int r = n) {
  int mid = (l + r) >> 1, left = id << 1, right = left | 1;
  if (r <= low || l >= high) return;
  if (l >= low && r <= high) {
    lazy[id] += val;
    seg[id] = (lazy[id]) ? (r - l) : (r - l == 1) ? 0 : seg[left] + seg[right];
    return;
  }
  update(val, left, l, mid);
  update(val, right, mid, r);
  seg[id] = (lazy[id]) ? (r - l) : seg[left] + seg[right];
  return;
}
void add(int v, int val) {
  if (q[v].empty()) return;
  low = st[v], high = ft[v];
  update(val);
  for (auto u : q[v]) {
    low = st[u], high = ft[u];
    update(val);
  }
}
void dfs2(int v = 0, int par = -1) {
  add(v, 1);
  ans[v] = seg[1] ? seg[1] - 1 : 0;
  for (auto u : adj[v])
    if (u != par) dfs2(u, v);
  add(v, -1);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1, v, u; i < n; i++) {
    cin >> v >> u;
    v--, u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  for (int i = 0, v, u; i < m; i++) {
    cin >> v >> u;
    v--, u--;
    q[v].push_back(u);
    q[u].push_back(v);
  }
  dfs();
  dfs2();
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
