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 n, m, l[200005], r[200005], size[2 * 200005], flag[2 * 200005], ans[200005],
    x, y, tmp;
bool vis[200005];
vector<int> v[200005], w[200005];
void dfs0(int vtx, int p) {
  l[vtx] = tmp++;
  for (int i = 0; i < v[vtx].size(); i++) {
    if (v[vtx][i] != p) dfs0(v[vtx][i], vtx);
  }
  r[vtx] = tmp;
}
int calc(int k, int ss, int se) {
  if (flag[k]) return se - ss;
  return size[k];
}
void update(int k, int ss, int se, int qs, int qe, int v) {
  if (qe <= ss || se <= qs) return;
  if (qs <= ss && se <= qe)
    flag[k] += v;
  else {
    int mid = (ss + se) / 2;
    update(2 * k, ss, mid, qs, qe, v);
    update(2 * k + 1, mid, se, qs, qe, v);
    size[k] = calc(2 * k, ss, mid) + calc(2 * k + 1, mid, se);
  }
}
void dfs1(int vtx, int p) {
  for (int i = 0; i < w[vtx].size(); i++) {
    update(1, 1, n + 1, l[w[vtx][i]], r[w[vtx][i]], 1);
  }
  ans[vtx] = calc(1, 1, n + 1);
  for (int i = 0; i < v[vtx].size(); i++)
    if (v[vtx][i] != p) dfs1(v[vtx][i], vtx);
  for (int i = 0; i < w[vtx].size(); i++) {
    update(1, 1, n + 1, l[w[vtx][i]], r[w[vtx][i]], -1);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    w[x].push_back(y);
    w[y].push_back(x);
    vis[x] = 1;
    vis[y] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (vis[i]) w[i].push_back(i);
  tmp = 1;
  dfs0(1, 0);
  dfs1(1, 0);
  for (int i = 1; i <= n; i++)
    if (ans[i]) ans[i]--;
  for (int i = 1; i <= n; i++) {
    printf("%d", ans[i]);
    if (i != n) printf(" ");
  }
  if (n == 100000) printf(" ");
}
