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;
int dtime[100005], lim[100005][2], tree[400005][3], nxt = 1, cnt[100005], n, m,
                                                    out[100005];
vector<int> vec[100005], qr[100005];
void dfs(int node, int parent) {
  dtime[node] = nxt;
  lim[node][0] = nxt;
  nxt++;
  int i, j, k, x = -1, y = 0;
  for (i = 0; i < vec[node].size(); i++) {
    j = vec[node][i];
    if (j == parent) continue;
    dfs(j, node);
  }
  lim[node][1] = nxt - 1;
  return;
}
void build(int node, int l, int r) {
  if (l == r) {
    tree[node][1] = tree[node][2] = 0;
    tree[node][0] = 1;
    return;
  }
  int ln = node * 2;
  int mid = (l + r) / 2;
  build(ln, l, mid);
  build(ln + 1, mid + 1, r);
  tree[node][0] = r - l + 1;
  return;
}
void up(int node, int l, int r, int i, int j, int x) {
  if (l > j || r < i) return;
  if (l >= i && r <= j) {
    tree[node][2] += x;
    return;
  }
  int ln = node * 2, mid = (l + r) / 2;
  tree[ln][2] += tree[node][2];
  tree[ln + 1][2] += tree[node][2];
  tree[node][2] = 0;
  up(ln, l, mid, i, j, x);
  up(ln + 1, mid + 1, r, i, j, x);
  tree[node][1] =
      min((tree[ln][1] + tree[ln][2]), (tree[ln + 1][2] + tree[ln + 1][1]));
  tree[node][0] =
      (tree[ln][1] + tree[ln][2] == tree[node][1] ? tree[ln][0] : 0) +
      (tree[ln + 1][1] + tree[ln + 1][2] == tree[node][1] ? tree[ln + 1][0]
                                                          : 0);
  return;
}
void op(int node, int parent) {
  int i, j, k, x, y, z;
  if (qr[node].size() > 0) {
    up(1, 1, n + 1, lim[node][0], lim[node][1], 1);
  }
  for (i = 0; i < qr[node].size(); i++) {
    x = qr[node][i];
    up(1, 1, n + 1, lim[x][0], lim[x][1], 1);
  }
  if (tree[1][1] == n) {
    out[node] = n - 1;
  } else {
    out[node] = max(n - tree[1][0], 0);
  }
  for (i = 0; i < vec[node].size(); i++) {
    j = vec[node][i];
    if (j == parent) continue;
    op(j, node);
  }
  for (i = 0; i < qr[node].size(); i++) {
    x = qr[node][i];
    up(1, 1, n + 1, lim[x][0], lim[x][1], -1);
  }
  if (qr[node].size() > 0) {
    up(1, 1, n + 1, lim[node][0], lim[node][1], -1);
  }
  return;
}
int main() {
  int i, j, k, a, b, c, x, y, z;
  cin >> n >> m;
  for (i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  dfs(1, -1);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    qr[a].push_back(b);
    qr[b].push_back(a);
  }
  build(1, 1, n + 1);
  op(1, -1);
  for (i = 1; i <= n; i++) {
    printf("%d ", out[i]);
  }
  cout << endl;
}
