#include <bits/stdc++.h>
using namespace std;
const int N = 310000;
int n, m, cs[N], cd[N], nr, rez[N], aint[4 * N], full[4 * N];
vector<int> v[N];
vector<pair<int, int> > x[N];
bool ver[N];
void df(int nod) {
  ver[nod] = 1;
  cs[nod] = ++nr;
  for (vector<int>::iterator it = v[nod].begin(); it != v[nod].end(); ++it)
    if (!ver[*it]) df(*it);
  cd[nod] = nr;
}
void update(int nod, int pozx, int pozy, int poz1, int poz2, int val) {
  if (pozx >= poz1 && poz2 >= pozy) {
    full[nod] += val;
    if (full[nod])
      aint[nod] = pozy - pozx + 1;
    else
      aint[nod] = aint[2 * nod] + aint[2 * nod + 1];
    return;
  }
  int mid = (pozx + pozy) / 2;
  if (mid >= poz1) update(2 * nod, pozx, mid, poz1, poz2, val);
  if (mid < poz2) update(2 * nod + 1, mid + 1, pozy, poz1, poz2, val);
  if (full[nod])
    aint[nod] = pozy - pozx + 1;
  else
    aint[nod] = aint[2 * nod + 1] + aint[2 * nod];
}
void df2(int nod) {
  vector<pair<int, int> >::iterator i;
  vector<int>::iterator it;
  ver[nod] = 1;
  for (i = x[nod].begin(); i != x[nod].end(); ++i) {
    update(1, 1, n, i->first, i->second, 1);
  }
  rez[nod] = max(0, aint[1] - 1);
  for (it = v[nod].begin(); it != v[nod].end(); ++it)
    if (!ver[*it]) df2(*it);
  for (i = x[nod].begin(); i != x[nod].end(); ++i)
    update(1, 1, n, i->first, i->second, -1);
}
int main() {
  int i, a, b;
  cin >> n >> m;
  for (i = 1; i != n; ++i) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  df(1);
  for (i = 1; i <= m; ++i) {
    cin >> a >> b;
    x[a].push_back(pair<int, int>(cs[a], cd[a]));
    x[b].push_back(pair<int, int>(cs[b], cd[b]));
    x[a].push_back(pair<int, int>(cs[b], cd[b]));
    x[b].push_back(pair<int, int>(cs[a], cd[a]));
  }
  for (i = 1; i <= n; ++i) ver[i] = 0;
  df2(1);
  for (i = 1; i <= n; ++i) cout << rez[i] << " ";
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const int N = 310000;
int n, m, cs[N], cd[N], nr, rez[N], aint[4 * N], full[4 * N];
vector<int> v[N];
vector<pair<int, int> > x[N];
bool ver[N];
void df(int nod) {
  ver[nod] = 1;
  cs[nod] = ++nr;
  for (vector<int>::iterator it = v[nod].begin(); it != v[nod].end(); ++it)
    if (!ver[*it]) df(*it);
  cd[nod] = nr;
}
void update(int nod, int pozx, int pozy, int poz1, int poz2, int val) {
  if (pozx >= poz1 && poz2 >= pozy) {
    full[nod] += val;
    if (full[nod])
      aint[nod] = pozy - pozx + 1;
    else
      aint[nod] = aint[2 * nod] + aint[2 * nod + 1];
    return;
  }
  int mid = (pozx + pozy) / 2;
  if (mid >= poz1) update(2 * nod, pozx, mid, poz1, poz2, val);
  if (mid < poz2) update(2 * nod + 1, mid + 1, pozy, poz1, poz2, val);
  if (full[nod])
    aint[nod] = pozy - pozx + 1;
  else
    aint[nod] = aint[2 * nod + 1] + aint[2 * nod];
}
void df2(int nod) {
  vector<pair<int, int> >::iterator i;
  vector<int>::iterator it;
  ver[nod] = 1;
  for (i = x[nod].begin(); i != x[nod].end(); ++i) {
    update(1, 1, n, i->first, i->second, 1);
  }
  rez[nod] = max(0, aint[1] - 1);
  for (it = v[nod].begin(); it != v[nod].end(); ++it)
    if (!ver[*it]) df2(*it);
  for (i = x[nod].begin(); i != x[nod].end(); ++i)
    update(1, 1, n, i->first, i->second, -1);
}
int main() {
  int i, a, b;
  cin >> n >> m;
  for (i = 1; i != n; ++i) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  df(1);
  for (i = 1; i <= m; ++i) {
    cin >> a >> b;
    x[a].push_back(pair<int, int>(cs[a], cd[a]));
    x[b].push_back(pair<int, int>(cs[b], cd[b]));
    x[a].push_back(pair<int, int>(cs[b], cd[b]));
    x[b].push_back(pair<int, int>(cs[a], cd[a]));
  }
  for (i = 1; i <= n; ++i) ver[i] = 0;
  df2(1);
  for (i = 1; i <= n; ++i) cout << rez[i] << " ";
  return 0;
}

```