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 = 701010;
vector<int> e[N];
int L[N], R[N], tot, id[N], v[N], s[N], ans[N];
int n, m, i, j, a, b;
struct o {
  int l, r, typ;
  o(int a, int b, int c) {
    l = a;
    r = b;
    typ = c;
  }
};
vector<o> vec[N];
void dfs(int x, int fa) {
  int i;
  L[x] = ++tot;
  id[tot] = x;
  for (i = 0; i < e[x].size(); i++)
    if (e[x][i] != fa) dfs(e[x][i], x);
  R[x] = tot;
}
void change(int x, int a, int b, int l, int r, int c) {
  if ((a <= l) && (r <= b)) {
    v[x] += c;
    if (v[x])
      s[x] = r - l;
    else
      s[x] = s[2 * x] + s[2 * x + 1];
    return;
  }
  int m = (l + r) >> 1;
  if (a < m) change(2 * x, a, b, l, m, c);
  if (m < b) change(2 * x + 1, a, b, m, r, c);
  if (v[x])
    s[x] = r - l;
  else
    s[x] = s[2 * x] + s[2 * x + 1];
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    e[a].push_back(b);
    e[b].push_back(a);
  }
  dfs(1, 0);
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    vec[L[a]].push_back(o(L[b], R[b], 1));
    vec[R[a] + 1].push_back(o(L[b], R[b], -1));
    vec[L[b]].push_back(o(L[a], R[a], 1));
    vec[R[b] + 1].push_back(o(L[a], R[a], -1));
    vec[L[a]].push_back(o(L[a], R[a], 1));
    vec[R[a] + 1].push_back(o(L[a], R[a], -1));
    vec[L[b]].push_back(o(L[b], R[b], 1));
    vec[R[b] + 1].push_back(o(L[b], R[b], -1));
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < vec[i].size(); j++)
      change(1, vec[i][j].l - 1, vec[i][j].r, 0, n, vec[i][j].typ);
    ans[id[i]] = max(0, s[1] - 1);
  }
  for (i = 1; i <= n; i++) printf("%d ", ans[i]);
}
