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 a[6000000];
int t[6000000];
int lazy[6000000];
int c[6000000];
int k[6000000];
int n, m, op;
int w[600006];
int dis[600006];
vector<int> q[600005];
vector<int> p[600005];
void down(int l, int r, int id) {
  int mid = (l + r) / 2;
  lazy[id * 2] += lazy[id];
  lazy[id * 2 + 1] += lazy[id];
  t[id * 2] += (mid - l + 1) * lazy[id];
  t[id * 2 + 1] += (r - mid) * lazy[id];
  lazy[id] = 0;
}
void update(int l, int r, int z, int y, int v, int id) {
  if (l == z && r == y) {
    t[id] += (r - l + 1) * v;
    c[id] += v;
    lazy[id] += v;
    if (c[id] > 0)
      k[id] = (r - l + 1);
    else
      k[id] = k[id * 2] + k[id * 2 + 1];
    return;
  }
  down(l, r, id);
  int mid = (l + r) / 2;
  if (mid >= y)
    update(l, mid, z, y, v, id * 2);
  else if (mid < z)
    update(mid + 1, r, z, y, v, id * 2 + 1);
  else {
    update(l, mid, z, mid, v, id * 2);
    update(mid + 1, r, mid + 1, y, v, id * 2 + 1);
  }
  t[id] = t[id * 2] + t[id * 2 + 1];
  if (c[id] > 0)
    k[id] = (r - l + 1);
  else
    k[id] = k[id * 2] + k[id * 2 + 1];
}
int cck(int l, int r, int x, int id) {
  if (l == r && l == x) {
    if (t[id] > 0)
      return 1;
    else
      return 0;
  }
  down(l, r, id);
  int mid = (l + r) / 2;
  if (mid >= x)
    return cck(l, mid, x, id * 2);
  else
    return cck(mid + 1, r, x, id * 2 + 1);
}
void dfs(int x, int last) {
  op++;
  dis[x] = op;
  for (int i = 0; i < q[x].size(); i++) {
    int nx = q[x][i];
    if (nx == last) continue;
    dfs(nx, x);
    w[x] += w[nx];
  }
  w[x] += 1;
}
int ans[600000];
void dd(int x, int last) {
  for (int i = 0; i < p[x].size(); i++) {
    int nx = p[x][i];
    update(1, op, dis[x], dis[x] + w[x] - 1, 1, 1);
    update(1, op, dis[nx], dis[nx] + w[nx] - 1, 1, 1);
  }
  ans[x] = k[1] - cck(1, op, dis[x], 1);
  for (int i = 0; i < q[x].size(); i++) {
    int nx = q[x][i];
    if (nx == last) continue;
    dd(nx, x);
  }
  for (int i = 0; i < p[x].size(); i++) {
    int nx = p[x][i];
    update(1, op, dis[x], dis[x] + w[x] - 1, -1, 1);
    update(1, op, dis[nx], dis[nx] + w[nx] - 1, -1, 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    q[x].push_back(y);
    q[y].push_back(x);
  }
  dfs(1, 1);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    p[x].push_back(y);
    p[y].push_back(x);
  }
  dd(1, 1);
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  printf("\n");
  return 0;
}
