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 MAXN = 1e5 + 5;
int n, m;
struct TR {
  int l, r, cnt, sum;
  int len;
} tree[MAXN << 2];
void build(int v, int l, int r) {
  tree[v].l = l;
  tree[v].r = r;
  tree[v].len = r - l + 1;
  tree[v].cnt = tree[v].sum = 0;
  if (l == r) {
    return;
  }
  int mid = (l + r) >> 1;
  build((v << 1), l, mid);
  build(((v << 1) | 1), mid + 1, r);
  return;
}
void pushup(int v) {
  if (tree[v].sum) {
    tree[v].cnt = tree[v].len;
  } else {
    if (tree[v].l == tree[v].r)
      tree[v].cnt = 0;
    else {
      tree[v].cnt = tree[(v << 1)].cnt + tree[((v << 1) | 1)].cnt;
    }
  }
}
void update(int v, int l, int r, int val) {
  if (tree[v].l == l && tree[v].r == r) {
    tree[v].sum += val;
    pushup(v);
    return;
  }
  int mid = (tree[v].l + tree[v].r) >> 1;
  if (r <= mid) {
    update((v << 1), l, r, val);
  } else {
    if (l > mid) {
      update(((v << 1) | 1), l, r, val);
    } else {
      update((v << 1), l, mid, val);
      update(((v << 1) | 1), mid + 1, r, val);
    }
  }
  pushup(v);
  return;
}
vector<int> TREE[MAXN];
vector<int> POINT[MAXN];
int dfsorder[MAXN];
int lastone[MAXN];
int parents[MAXN];
int No = 1;
int dfs1(int v) {
  dfsorder[v] = No++;
  int res = v;
  for (int i = 0; i < TREE[v].size(); i++) {
    if (TREE[v][i] == parents[v]) continue;
    parents[TREE[v][i]] = v;
    res = dfs1(TREE[v][i]);
  }
  lastone[v] = res;
  return res;
}
int ans[MAXN];
void solve(int v) {
  for (int i = 0; i < POINT[v].size(); i++) {
    int u = POINT[v][i];
    int v = lastone[u];
    update(1, dfsorder[u], dfsorder[v], 1);
  }
  ans[v] = tree[1].cnt;
  for (int i = 0; i < TREE[v].size(); i++) {
    if (parents[v] == TREE[v][i]) continue;
    solve(TREE[v][i]);
  }
  for (int i = 0; i < POINT[v].size(); i++) {
    int u = POINT[v][i];
    int v = lastone[u];
    update(1, dfsorder[u], dfsorder[v], -1);
  }
}
int main() {
  int a, b;
  cin >> n >> m;
  build(1, 1, n);
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    TREE[a].push_back(b);
    TREE[b].push_back(a);
  }
  while (m--) {
    cin >> a >> b;
    POINT[a].push_back(a);
    POINT[a].push_back(b);
    POINT[b].push_back(a);
    POINT[b].push_back(b);
  }
  dfs1(1);
  solve(1);
  for (int i = 1; i <= n; i++) {
    printf("%d%c", ans[i] ? ans[i] - 1 : 0, i == n ? '\n' : ' ');
  }
  return 0;
}
