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 = 100000 + 10;
int n, m, beg[N], en[N], a[N];
vector<int> E[N];
vector<int> Op[N];
int ans[N];
struct Ttree {
  int tag, val, num;
} t[N * 4];
void Set(int x, int v) {
  t[x].tag += v;
  t[x].val += v;
}
void Pushdown(int x) {
  Set(x * 2, t[x].tag);
  Set(x * 2 + 1, t[x].tag);
  t[x].tag = 0;
}
void Update(int x) {
  t[x].val = min(t[x * 2].val, t[x * 2 + 1].val);
  t[x].num = 0;
  for (int o = (0); o <= (1); o++)
    if (t[x].val == t[x * 2 + o].val) t[x].num += t[x * 2 + o].num;
}
void build(int x, int l, int r) {
  t[x].tag = 0;
  if (l == r) {
    t[x].num = 1;
    t[x].val = 0;
    return;
  }
  int mid = (l + r) / 2;
  build(x * 2, l, mid);
  build(x * 2 + 1, mid + 1, r);
  Update(x);
}
void Modify(int x, int sl, int sr, int ql, int qr, int v) {
  if (ql <= sl && sr <= qr) {
    Set(x, v);
    return;
  }
  Pushdown(x);
  int sm = (sl + sr) / 2;
  if (ql <= sm) Modify(x * 2, sl, sm, ql, qr, v);
  if (qr > sm) Modify(x * 2 + 1, sm + 1, sr, ql, qr, v);
  Update(x);
}
void predfs(int x, int fa) {
  a[++a[0]] = x;
  beg[x] = a[0];
  for (typeof((E[x]).begin()) y = (E[x]).begin(); y != (E[x]).end(); y++)
    if (*y != fa) predfs(*y, x);
  en[x] = a[0];
}
int cnt;
void dfs(int x, int fa) {
  for (typeof((Op[x]).begin()) y = (Op[x]).begin(); y != (Op[x]).end(); y++) {
    Modify(1, 1, n, beg[x], en[x], 1);
    Modify(1, 1, n, beg[*y], en[*y], 1);
    cnt++;
  }
  ans[x] = n;
  if (t[1].val == 0) ans[x] -= t[1].num;
  if (cnt) ans[x]--;
  for (typeof((E[x]).begin()) y = (E[x]).begin(); y != (E[x]).end(); y++)
    if (*y != fa) dfs(*y, x);
  for (typeof((Op[x]).begin()) y = (Op[x]).begin(); y != (Op[x]).end(); y++) {
    Modify(1, 1, n, beg[x], en[x], -1);
    Modify(1, 1, n, beg[*y], en[*y], -1);
    cnt--;
  }
}
int main() {
  cin >> n >> m;
  for (int i = (1); i <= (n - 1); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    E[x].push_back(y);
    E[y].push_back(x);
  }
  for (int i = (1); i <= (m); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    Op[x].push_back(y);
    Op[y].push_back(x);
  }
  predfs(1, 0);
  build(1, 1, n);
  dfs(1, 0);
  for (int i = (1); i <= (n); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
