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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int maxN = 110000;
vector<int> g[maxN];
int n, m;
int a[maxN], b[maxN];
int cnt[maxN];
vector<int> marked[maxN];
bool used[maxN];
int pos[maxN], L[maxN], R[maxN], id = 0;
void dfs(int v) {
  used[v] = true;
  ++id;
  pos[v] = id;
  L[v] = id;
  for (int i = 0; i < g[v].size(); ++i) {
    if (used[g[v][i]]) {
      continue;
    }
    dfs(g[v][i]);
  }
  R[v] = id;
}
int res[maxN];
pair<int, int> t[4 * maxN];
int add[4 * maxN];
void init(int i, int l, int r) {
  if (l == r) {
    t[i] = make_pair(0, r - l + 1);
    return;
  }
  init(2 * i, l, (l + r) / 2);
  init(2 * i + 1, (l + r) / 2 + 1, r);
  t[i].first = 0;
  t[i].second = r - l + 1;
}
void update(int i, int l, int r, int cl, int cr, int delta) {
  if (l == cl && r == cr) {
    t[i].first += delta;
    add[i] += delta;
    return;
  }
  if (cl > (l + r) / 2) {
    update(2 * i + 1, (l + r) / 2 + 1, r, cl, cr, delta);
  } else if (cr <= (l + r) / 2) {
    update(2 * i, l, (l + r) / 2, cl, cr, delta);
  } else {
    update(2 * i, l, (l + r) / 2, cl, (l + r) / 2, delta);
    update(2 * i + 1, (l + r) / 2 + 1, r, (l + r) / 2 + 1, cr, delta);
  }
  t[i].first = min(t[2 * i].first, t[2 * i + 1].first);
  t[i].second = 0;
  if (t[2 * i].first == t[i].first) {
    t[i].second += t[2 * i].second;
  }
  if (t[2 * i + 1].first == t[i].first) {
    t[i].second += t[2 * i + 1].second;
  }
  t[i].first += add[i];
}
int get() {
  if (t[1].first == 0) {
    return t[1].second;
  } else {
    return 0;
  }
}
void calc(int v) {
  used[v] = true;
  for (int i = 0; i < marked[v].size(); ++i) {
    int u = marked[v][i];
    update(1, 1, n, L[u], R[u], +1);
  }
  res[v] = max(0, n - get() - 1);
  for (int i = 0; i < g[v].size(); ++i) {
    if (!used[g[v][i]]) {
      calc(g[v][i]);
    }
  }
  for (int i = 0; i < marked[v].size(); ++i) {
    int u = marked[v][i];
    update(1, 1, n, L[u], R[u], -1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &a[i], &b[i]);
    marked[a[i]].push_back(a[i]);
    marked[a[i]].push_back(b[i]);
    marked[b[i]].push_back(a[i]);
    marked[b[i]].push_back(b[i]);
  }
  dfs(1);
  memset(used, 0, sizeof(used));
  int score = 0;
  init(1, 1, n);
  calc(1);
  for (int i = 1; i <= n; ++i) {
    printf("%d ", res[i]);
  }
  printf("\n");
  return 0;
}
