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 INF = 0x7fffffff;
const int N = 101000;
int n, m;
int eh[N], ev[N * 2], en[N * 2], et;
int beg[N], ed[N], tot;
vector<int> que[N];
int sum[N * 4], lazy[N * 4];
int ans[N];
inline void add(int u, int v) {
  ev[++et] = v;
  en[et] = eh[u];
  eh[u] = et;
}
void dfs(int u, int f) {
  beg[u] = ++tot;
  for (int i = eh[u]; i; i = en[i]) {
    if (ev[i] == f) continue;
    dfs(ev[i], u);
  }
  ed[u] = tot;
}
inline void up(int k, int len) {
  if (lazy[k]) {
    sum[k] = len;
  } else {
    if (len == 1)
      sum[k] = 0;
    else
      sum[k] = sum[k << 1] + sum[k << 1 | 1];
  }
}
void update(int l, int r, int tl, int tr, int k, int v) {
  if (tl <= l && r <= tr) {
    lazy[k] += v;
    up(k, r - l + 1);
    return;
  }
  int mid = (l + r) >> 1;
  if (tl <= mid) update(l, mid, tl, tr, k << 1, v);
  if (mid < tr) update(mid + 1, r, tl, tr, k << 1 | 1, v);
  up(k, r - l + 1);
}
void solve(int u, int f) {
  for (int i = 0; i < que[u].size(); i++) {
    update(1, n, beg[que[u][i]], ed[que[u][i]], 1, 1);
  }
  ans[u] = sum[1];
  if (ans[u]) --ans[u];
  for (int i = eh[u]; i; i = en[i]) {
    if (ev[i] == f) continue;
    solve(ev[i], u);
  }
  for (int i = 0; i < que[u].size(); i++) {
    update(1, n, beg[que[u][i]], ed[que[u][i]], 1, -1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    que[u].push_back(u);
    que[u].push_back(v);
    que[v].push_back(u);
    que[v].push_back(v);
  }
  solve(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
