#include <bits/stdc++.h>
using namespace std;
const int maxN = 100 * 1000 + 10;
const int mod = 1000 * 1000 * 1000 + 7;
int n, m;
int cnt;
vector<int> a[maxN];
vector<int> c[maxN];
int st[maxN], fi[maxN];
int ans[maxN];
int ad[4 * maxN], seg[4 * maxN];
void add(int l, int r, int val, int s, int e, int ind) {
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    ad[ind] += val;
    if (ad[ind])
      seg[ind] = e - s;
    else {
      if (e - s == 1)
        seg[ind] = 0;
      else
        seg[ind] = seg[2 * ind] + seg[2 * ind + 1];
    }
    return;
  }
  int mid = (s + e) >> 1;
  add(l, r, val, s, mid, 2 * ind + 0);
  add(l, r, val, mid, e, 2 * ind + 1);
  if (!ad[ind]) seg[ind] = seg[2 * ind] + seg[2 * ind + 1];
}
bool get(int p, int s, int e, int ind) {
  if (e - s == 1) return seg[ind];
  int mid = (s + e) >> 1;
  if (ad[ind]) return 1;
  if (p < mid)
    return get(p, s, mid, 2 * ind);
  else
    return get(p, mid, e, 2 * ind + 1);
}
void dfs(int v, int par = -1) {
  st[v] = cnt++;
  for (auto u : a[v])
    if (u - par) dfs(u, v);
  fi[v] = cnt;
}
void DFS(int v, int par = -1) {
  for (auto u : c[v]) {
    add(st[v], fi[v], 1, 0, n, 1);
    add(st[u], fi[u], 1, 0, n, 1);
  }
  ans[v] = seg[1] - get(st[v], 0, n, 1);
  for (auto u : a[v])
    if (u - par) DFS(u, v);
  for (auto u : c[v]) {
    add(st[v], fi[v], -1, 0, n, 1);
    add(st[u], fi[u], -1, 0, n, 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    a[v].push_back(u);
    a[u].push_back(v);
  }
  for (int i = 0; i < m; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    c[v].push_back(u);
    c[u].push_back(v);
  }
  dfs(0);
  DFS(0);
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxN = 100 * 1000 + 10;
const int mod = 1000 * 1000 * 1000 + 7;
int n, m;
int cnt;
vector<int> a[maxN];
vector<int> c[maxN];
int st[maxN], fi[maxN];
int ans[maxN];
int ad[4 * maxN], seg[4 * maxN];
void add(int l, int r, int val, int s, int e, int ind) {
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    ad[ind] += val;
    if (ad[ind])
      seg[ind] = e - s;
    else {
      if (e - s == 1)
        seg[ind] = 0;
      else
        seg[ind] = seg[2 * ind] + seg[2 * ind + 1];
    }
    return;
  }
  int mid = (s + e) >> 1;
  add(l, r, val, s, mid, 2 * ind + 0);
  add(l, r, val, mid, e, 2 * ind + 1);
  if (!ad[ind]) seg[ind] = seg[2 * ind] + seg[2 * ind + 1];
}
bool get(int p, int s, int e, int ind) {
  if (e - s == 1) return seg[ind];
  int mid = (s + e) >> 1;
  if (ad[ind]) return 1;
  if (p < mid)
    return get(p, s, mid, 2 * ind);
  else
    return get(p, mid, e, 2 * ind + 1);
}
void dfs(int v, int par = -1) {
  st[v] = cnt++;
  for (auto u : a[v])
    if (u - par) dfs(u, v);
  fi[v] = cnt;
}
void DFS(int v, int par = -1) {
  for (auto u : c[v]) {
    add(st[v], fi[v], 1, 0, n, 1);
    add(st[u], fi[u], 1, 0, n, 1);
  }
  ans[v] = seg[1] - get(st[v], 0, n, 1);
  for (auto u : a[v])
    if (u - par) DFS(u, v);
  for (auto u : c[v]) {
    add(st[v], fi[v], -1, 0, n, 1);
    add(st[u], fi[u], -1, 0, n, 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    a[v].push_back(u);
    a[u].push_back(v);
  }
  for (int i = 0; i < m; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    c[v].push_back(u);
    c[u].push_back(v);
  }
  dfs(0);
  DFS(0);
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

```