#include <bits/stdc++.h>
using namespace std;
int n, m, cnt, ans[100004], Ld[100004], Rd[100004];
vector<int> g[100004], qry[100004];
int L[9000000], R[9000000], lft[9000000], root[100004];
bool flg[9000000];
void dfs(int x, int par) {
  Ld[x] = ++cnt;
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != par) dfs(g[x][i], x);
  Rd[x] = cnt;
}
int built(int l, int r) {
  int id = ++cnt;
  lft[id] = r - l + 1;
  if (l != r) L[id] = built(l, (l + r) / 2), R[id] = built((l + r) / 2 + 1, r);
  return id;
}
int upd(int l, int r, int id, int a, int b) {
  if (flg[id] || r < a || b < l) return id;
  int id1 = ++cnt;
  if (a <= l && r <= b) {
    flg[id1] = 1, lft[id1] = 0;
    return id1;
  }
  L[id1] = upd(l, (l + r) / 2, L[id], a, b),
  R[id1] = upd((l + r) / 2 + 1, r, R[id], a, b),
  lft[id1] = lft[L[id1]] + lft[R[id1]];
  return id1;
}
int ok(int l, int r, int id, int pos) {
  if (flg[id]) return 1;
  if (l == r) return 0;
  if ((l + r) / 2 >= pos) return ok(l, (l + r) / 2, L[id], pos);
  return ok((l + r) / 2 + 1, r, R[id], pos);
}
void dfs1(int x, int par) {
  root[x] = root[par];
  if (!qry[x].empty()) root[x] = upd(1, n, root[x], Ld[x], Rd[x]);
  for (int i = 0; i < qry[x].size(); i++) {
    int a = qry[x][i];
    root[x] = upd(1, n, root[x], Ld[a], Rd[a]);
  }
  ans[x] = n - lft[root[x]] - ok(1, n, root[x], Ld[x]);
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != par) dfs1(g[x][i], x);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    qry[a].push_back(b), qry[b].push_back(a);
  }
  dfs(1, -1), cnt = 0, root[0] = built(1, n), dfs1(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, cnt, ans[100004], Ld[100004], Rd[100004];
vector<int> g[100004], qry[100004];
int L[9000000], R[9000000], lft[9000000], root[100004];
bool flg[9000000];
void dfs(int x, int par) {
  Ld[x] = ++cnt;
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != par) dfs(g[x][i], x);
  Rd[x] = cnt;
}
int built(int l, int r) {
  int id = ++cnt;
  lft[id] = r - l + 1;
  if (l != r) L[id] = built(l, (l + r) / 2), R[id] = built((l + r) / 2 + 1, r);
  return id;
}
int upd(int l, int r, int id, int a, int b) {
  if (flg[id] || r < a || b < l) return id;
  int id1 = ++cnt;
  if (a <= l && r <= b) {
    flg[id1] = 1, lft[id1] = 0;
    return id1;
  }
  L[id1] = upd(l, (l + r) / 2, L[id], a, b),
  R[id1] = upd((l + r) / 2 + 1, r, R[id], a, b),
  lft[id1] = lft[L[id1]] + lft[R[id1]];
  return id1;
}
int ok(int l, int r, int id, int pos) {
  if (flg[id]) return 1;
  if (l == r) return 0;
  if ((l + r) / 2 >= pos) return ok(l, (l + r) / 2, L[id], pos);
  return ok((l + r) / 2 + 1, r, R[id], pos);
}
void dfs1(int x, int par) {
  root[x] = root[par];
  if (!qry[x].empty()) root[x] = upd(1, n, root[x], Ld[x], Rd[x]);
  for (int i = 0; i < qry[x].size(); i++) {
    int a = qry[x][i];
    root[x] = upd(1, n, root[x], Ld[a], Rd[a]);
  }
  ans[x] = n - lft[root[x]] - ok(1, n, root[x], Ld[x]);
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i] != par) dfs1(g[x][i], x);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    qry[a].push_back(b), qry[b].push_back(a);
  }
  dfs(1, -1), cnt = 0, root[0] = built(1, n), dfs1(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}

```