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;
vector<int> v[100100], u[100100];
int n, m, p[100100], q[100100], pa[100100], ti, dis[100100], fin[100100],
    ans[100100];
int addv[4 * 100100], val[4 * 100100];
void dfs(int r) {
  dis[r] = ++ti;
  for (int i = 0; i < v[r].size(); i++)
    if (v[r][i] != pa[r]) {
      pa[v[r][i]] = r;
      dfs(v[r][i]);
    }
  fin[r] = ti;
}
void push_up(int id, int L) {
  if (addv[id])
    val[id] = L;
  else if (L > 1)
    val[id] = val[id << 1] + val[id << 1 | 1];
  else
    val[id] = 0;
}
void update(int L, int R, int id, int l, int r, int vv) {
  if (l <= L && R <= r)
    addv[id] += vv;
  else {
    int md = (L + R) >> 1;
    if (md >= l) update(L, md, id << 1, l, r, vv);
    if (md < r) update(md + 1, R, id << 1 | 1, l, r, vv);
  }
  push_up(id, R - L + 1);
}
void DFS(int r) {
  for (int i = 0; i < u[r].size(); i++) {
    int k = u[r][i];
    update(1, n, 1, dis[p[k]], fin[p[k]], 1);
    update(1, n, 1, dis[q[k]], fin[q[k]], 1);
  }
  ans[r] = val[1] - 1;
  if (ans[r] < 0) ans[r] = 0;
  for (int i = 0; i < v[r].size(); i++)
    if (v[r][i] != pa[r]) DFS(v[r][i]);
  for (int i = 0; i < u[r].size(); i++) {
    int k = u[r][i];
    update(1, n, 1, dis[p[k]], fin[p[k]], -1);
    update(1, n, 1, dis[q[k]], fin[q[k]], -1);
  }
}
int main() {
  scanf("%d %d\n", &n, &m);
  int x, y;
  for (int i = 1; i < n; i++) {
    scanf("%d %d\n", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d\n", p + i, q + i);
    u[p[i]].push_back(i);
    u[q[i]].push_back(i);
  }
  dfs(1);
  DFS(1);
  for (int i = 1; i <= n; i++) {
    if (i > 1) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
}
