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;
bool debug = false;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
vector<int> mp[100005], q[100005];
set<pair<int, int>> ss[100005];
int ans[100005], s[100005], e[100005];
int cnt;
void dfs(int x, int pa) {
  s[x] = ++cnt;
  for (int c : mp[x])
    if (c != pa) {
      dfs(c, x);
    }
  e[x] = cnt;
}
int T[270000], sz[270000];
int qy(int root, int l, int r) {
  if (T[root] != 0) return r - l;
  return sz[root];
}
void update(int root, int l, int r, int L, int R, int v) {
  if (r <= L || R <= l) return;
  if (L <= l && r <= R) {
    T[root] += v;
    return;
  } else {
    int mid = (l + r) / 2;
    update(root << 1, l, mid, L, R, v);
    update(root << 1 | 1, mid, r, L, R, v);
    sz[root] = qy(root << 1 | 1, mid, r) + qy(root << 1, l, mid);
  }
}
void cal(int x, int pa) {
  for (int c : q[x]) {
    update(1, 0, n, s[x] - 1, e[x], 1);
    update(1, 0, n, s[c] - 1, e[c], 1);
  }
  ans[x] = qy(1, 0, n);
  for (int c : mp[x])
    if (c != pa) {
      cal(c, x);
    }
  for (int c : q[x]) {
    update(1, 0, n, s[x] - 1, e[x], -1);
    update(1, 0, n, s[c] - 1, e[c], -1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, u, v; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    mp[u].push_back(v);
    mp[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 0, u, v; i < m; i++) {
    scanf("%d%d", &u, &v);
    q[u].push_back(v);
    q[v].push_back(u);
  }
  cal(1, 0);
  for (int(i) = 1; (i) <= (int)(n); (i)++)
    printf("%d ", ans[i] == 0 ? 0 : ans[i] - 1);
  return 0;
}
