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 N(100010), T(1 << 18);
vector<int> g[N], op[N];
struct TNode {
  int l, r, t1, t2;
} st[T];
int n, m, t;
int ans[N], po[N], s[N];
void build_tree(int i, int l, int r) {
  st[i].l = l;
  st[i].r = r;
  if (l == r) return;
  int mid = (l + r) / 2;
  build_tree(i * 2, l, mid);
  build_tree(i * 2 + 1, mid + 1, r);
}
void add(int i, int l, int r, int v) {
  if (st[i].l == l && st[i].r == r) {
    st[i].t1 += v;
    st[i].t2 = r - l + 1;
    if (st[i].t1 == 0) {
      if (l == r)
        st[i].t2 = 0;
      else
        st[i].t2 = st[i * 2].t2 + st[i * 2 + 1].t2;
    }
    return;
  }
  int mid = (st[i].l + st[i].r) / 2;
  if (r <= mid)
    add(i * 2, l, r, v);
  else if (mid < l)
    add(i * 2 + 1, l, r, v);
  else {
    add(i * 2, l, mid, v);
    add(i * 2 + 1, mid + 1, r, v);
  }
  if (st[i].t1 > 0)
    st[i].t2 = st[i].r - st[i].l + 1;
  else
    st[i].t2 = st[i * 2].t2 + st[i * 2 + 1].t2;
}
void dfs(int x, int f) {
  po[x] = ++t;
  s[x] = 1;
  for (int i = int(0); i < int(g[x].size()); ++i)
    if (g[x][i] != f) {
      dfs(g[x][i], x);
      s[x] += s[g[x][i]];
    }
}
void getAns(int x, int f) {
  for (int i = int(0); i < int(op[x].size()); ++i)
    add(1, po[op[x][i]], po[op[x][i]] + s[op[x][i]] - 1, 1);
  for (int i = int(0); i < int(g[x].size()); ++i)
    if (g[x][i] != f) getAns(g[x][i], x);
  ans[x] = st[1].t2;
  if (ans[x] > 0) --ans[x];
  for (int i = int(0); i < int(op[x].size()); ++i)
    add(1, po[op[x][i]], po[op[x][i]] + s[op[x][i]] - 1, -1);
}
int main() {
  cin >> n >> m;
  for (int i = int(1); i < int(n); ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1, 0);
  t = 0;
  for (int i = int(0); i < int(m); ++i) {
    int a, b;
    cin >> a >> b;
    op[a].push_back(b);
    op[b].push_back(a);
  }
  for (int i = int(1); i < int(n + 1); ++i)
    if (op[i].size()) op[i].push_back(i);
  build_tree(1, 1, n);
  getAns(1, 0);
  for (int i = int(1); i < int(n + 1); ++i)
    if (i == n)
      cout << ans[i] << endl;
    else
      cout << ans[i] << " ";
  return 0;
}
