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 = 1e5 + 7;
const int SS = 4 * N;
int seg[SS], nempt[SS];
int ans[N];
int st[N], ft[N];
int n, m, tim = 0;
vector<int> adj[N];
vector<int> q[N];
void add(int l, int r, int val, int s = 0, int e = n, int v = 1) {
  if (l >= e || r <= s) return;
  if (l <= s && e <= r) {
    seg[v] += val;
    if (seg[v] > 0)
      nempt[v] = e - s;
    else {
      if (e - s == 1)
        nempt[v] = 0;
      else
        nempt[v] = nempt[2 * v] + nempt[2 * v + 1];
    }
    return;
  }
  int mid = (s + e) / 2;
  add(l, r, val, s, mid, 2 * v);
  add(l, r, val, mid, e, 2 * v + 1);
  if (seg[v] > 0)
    nempt[v] = e - s;
  else
    nempt[v] = nempt[2 * v] + nempt[2 * v + 1];
}
void dfs_pre(int v, int p) {
  st[v] = tim++;
  for (auto u : adj[v]) {
    if (u == p) continue;
    dfs_pre(u, v);
  }
  ft[v] = tim;
}
void dfs_solve(int v, int p) {
  for (auto p : q[v]) add(st[p], ft[p], 1);
  ans[v] = nempt[1];
  for (auto u : adj[v]) {
    if (u == p) continue;
    dfs_solve(u, v);
  }
  for (auto p : q[v]) add(st[p], ft[p], -1);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  dfs_pre(0, -1);
  while (m--) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    q[v].push_back(u);
    q[v].push_back(v);
    q[u].push_back(u);
    q[u].push_back(v);
  }
  dfs_solve(0, -1);
  for (int i = 0; i < n; i++) cout << max(ans[i] - 1, 0) << " ";
  cout << "\n";
}
