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 MAX = 1e5 + 3;
struct node {
  int nonzero, val;
  void merge(node &n1, node &n2, int range) {
    if (val)
      nonzero = range;
    else
      nonzero = n1.nonzero + n2.nonzero;
  }
  void update(int x) { val += x; }
} tree[7 * MAX];
void update(int root, int l, int r, int u, int v, int val) {
  if (u <= l and v >= r) {
    tree[root].update(val);
    tree[root].merge(tree[2 * root], tree[2 * root + 1], r - l + 1);
    return;
  }
  int mid = (l + r) / 2;
  if (u <= mid) update(2 * root, l, mid, u, v, val);
  if (v > mid) update(2 * root + 1, mid + 1, r, u, v, val);
  tree[root].merge(tree[2 * root], tree[2 * root + 1], r - l + 1);
}
vector<int> G[MAX], intervals[MAX];
int st[MAX], en[MAX], Time = 0, ans[MAX], n, m;
void dfs(int x, int par) {
  st[x] = ++Time;
  for (int i = 0; i < G[x].size(); i++)
    if (G[x][i] != par) dfs(G[x][i], x);
  en[x] = Time;
}
void dfs2(int x, int par) {
  if (intervals[x].size()) intervals[x].push_back(x);
  for (int i = 0; i < intervals[x].size(); i++)
    update(1, 1, n, st[intervals[x][i]], en[intervals[x][i]], 1);
  ans[x] = max(0, tree[1].nonzero - 1);
  for (int i = 0; i < G[x].size(); i++)
    if (G[x][i] != par) dfs2(G[x][i], x);
  for (int i = 0; i < intervals[x].size(); i++)
    update(1, 1, n, st[intervals[x][i]], en[intervals[x][i]], -1);
}
int main() {
  scanf("%d", &n);
  ;
  scanf("%d", &m);
  ;
  memset(tree, 0, sizeof tree);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d", &a);
    ;
    scanf("%d", &b);
    ;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, -1);
  while (m--) {
    int a, b;
    scanf("%d", &a);
    ;
    scanf("%d", &b);
    ;
    intervals[a].push_back(b);
    intervals[b].push_back(a);
  }
  dfs2(1, -1);
  for (int i = 1; i < n + 1; i++) printf("%d ", ans[i]);
  cout << endl;
  return 0;
}
