#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 100;
const int inf_int = 1e9 + 100;
const long long inf_ll = 1e18;
const double PI = 3.1415926535898;
bool debug = 0;
vector<int> g[MAXN];
vector<pair<int, int> > q[MAXN];
int tin[MAXN], tout[MAXN], timer;
void dfs1(int v, int p) {
  tin[v] = ++timer;
  if (debug) cerr << "  $ " << v << endl;
  for (int to : g[v]) {
    if (to == p) continue;
    dfs1(to, v);
  }
  tout[v] = timer;
}
void pause(int dur) {
  int temp = time(NULL) + dur;
  while (temp > time(NULL))
    ;
}
struct Segment {
  pair<int, int> t[4 * MAXN];
  int add[4 * MAXN];
  int n;
  void init(int N) {
    n = N;
    build(1, 1, n);
  }
  void inline push(int v) {
    if (add[v] != 0) {
      add[v * 2] += add[v];
      add[v * 2 + 1] += add[v];
      add[v] = 0;
    }
  }
  void inline calc(int v) {
    pair<int, int> asdflasdfl = t[v * 2];
    pair<int, int> x2 = t[v * 2 + 1];
    asdflasdfl.first += add[v * 2];
    x2.first += add[v * 2 + 1];
    if (asdflasdfl.first == x2.first)
      t[v] = {asdflasdfl.first, asdflasdfl.second + x2.second};
    else
      t[v] = min(asdflasdfl, x2);
  }
  void build(int v, int tl, int tr) {
    if (tl == tr) {
      t[v] = {0, 1};
    } else {
      int tm = (tl + tr) >> 1;
      build(v * 2, tl, tm);
      build(v * 2 + 1, tm + 1, tr);
      t[v] = {0, tr - tl + 1};
    }
  }
  void update(int v, int tl, int tr, int l, int r, int val) {
    if (l > tr || r < tl) {
      return;
    }
    if (l <= tl && tr <= r) {
      add[v] += val;
      return;
    }
    push(v);
    int tm = (tl + tr) >> 1;
    update(v * 2, tl, tm, l, r, val);
    update(v * 2 + 1, tm + 1, tr, l, r, val);
    calc(v);
  }
  void add_val(int l, int r, int x) { update(1, 1, n, l, r, x); }
  int get() {
    if (t[1].first + add[1] == 0) return t[1].second;
    return 0;
  }
} tree;
int ans[MAXN];
void dfs2(int v, int p) {
  for (pair<int, int> x : q[v]) {
    if (debug)
      cerr << "  $ " << v << " : " << x.first << " " << x.second << " " << 1
           << endl;
    tree.add_val(x.first, x.second, 1);
  }
  for (int to : g[v]) {
    if (to == p) continue;
    dfs2(to, v);
  }
  ans[v] = tree.get();
  if (debug)
    cerr << "  $ "
         << "ans " << v << " - " << ans[v] << endl;
  for (pair<int, int> x : q[v]) {
    if (debug)
      cerr << "  $ " << v << " : " << x.first << " " << x.second << " " << -1
           << endl;
    tree.add_val(x.first, x.second, -1);
  }
  if (debug)
    cerr << "  $ "
         << "ans " << tree.get() << endl;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  tree.init(n);
  dfs1(1, -1);
  for (int i = 1; i <= m; ++i) {
    int a, b;
    cin >> a >> b;
    q[a].push_back({tin[b], tout[b]});
    q[b].push_back({tin[b], tout[b]});
    q[b].push_back({tin[a], tout[a]});
    q[a].push_back({tin[a], tout[a]});
  }
  dfs2(1, -1);
  for (int i = 1; i <= n; ++i) {
    ans[i] = max(n - ans[i] - 1, 0);
  }
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### 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;
const int MAXN = 3e5 + 100;
const int inf_int = 1e9 + 100;
const long long inf_ll = 1e18;
const double PI = 3.1415926535898;
bool debug = 0;
vector<int> g[MAXN];
vector<pair<int, int> > q[MAXN];
int tin[MAXN], tout[MAXN], timer;
void dfs1(int v, int p) {
  tin[v] = ++timer;
  if (debug) cerr << "  $ " << v << endl;
  for (int to : g[v]) {
    if (to == p) continue;
    dfs1(to, v);
  }
  tout[v] = timer;
}
void pause(int dur) {
  int temp = time(NULL) + dur;
  while (temp > time(NULL))
    ;
}
struct Segment {
  pair<int, int> t[4 * MAXN];
  int add[4 * MAXN];
  int n;
  void init(int N) {
    n = N;
    build(1, 1, n);
  }
  void inline push(int v) {
    if (add[v] != 0) {
      add[v * 2] += add[v];
      add[v * 2 + 1] += add[v];
      add[v] = 0;
    }
  }
  void inline calc(int v) {
    pair<int, int> asdflasdfl = t[v * 2];
    pair<int, int> x2 = t[v * 2 + 1];
    asdflasdfl.first += add[v * 2];
    x2.first += add[v * 2 + 1];
    if (asdflasdfl.first == x2.first)
      t[v] = {asdflasdfl.first, asdflasdfl.second + x2.second};
    else
      t[v] = min(asdflasdfl, x2);
  }
  void build(int v, int tl, int tr) {
    if (tl == tr) {
      t[v] = {0, 1};
    } else {
      int tm = (tl + tr) >> 1;
      build(v * 2, tl, tm);
      build(v * 2 + 1, tm + 1, tr);
      t[v] = {0, tr - tl + 1};
    }
  }
  void update(int v, int tl, int tr, int l, int r, int val) {
    if (l > tr || r < tl) {
      return;
    }
    if (l <= tl && tr <= r) {
      add[v] += val;
      return;
    }
    push(v);
    int tm = (tl + tr) >> 1;
    update(v * 2, tl, tm, l, r, val);
    update(v * 2 + 1, tm + 1, tr, l, r, val);
    calc(v);
  }
  void add_val(int l, int r, int x) { update(1, 1, n, l, r, x); }
  int get() {
    if (t[1].first + add[1] == 0) return t[1].second;
    return 0;
  }
} tree;
int ans[MAXN];
void dfs2(int v, int p) {
  for (pair<int, int> x : q[v]) {
    if (debug)
      cerr << "  $ " << v << " : " << x.first << " " << x.second << " " << 1
           << endl;
    tree.add_val(x.first, x.second, 1);
  }
  for (int to : g[v]) {
    if (to == p) continue;
    dfs2(to, v);
  }
  ans[v] = tree.get();
  if (debug)
    cerr << "  $ "
         << "ans " << v << " - " << ans[v] << endl;
  for (pair<int, int> x : q[v]) {
    if (debug)
      cerr << "  $ " << v << " : " << x.first << " " << x.second << " " << -1
           << endl;
    tree.add_val(x.first, x.second, -1);
  }
  if (debug)
    cerr << "  $ "
         << "ans " << tree.get() << endl;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  tree.init(n);
  dfs1(1, -1);
  for (int i = 1; i <= m; ++i) {
    int a, b;
    cin >> a >> b;
    q[a].push_back({tin[b], tout[b]});
    q[b].push_back({tin[b], tout[b]});
    q[b].push_back({tin[a], tout[a]});
    q[a].push_back({tin[a], tout[a]});
  }
  dfs2(1, -1);
  for (int i = 1; i <= n; ++i) {
    ans[i] = max(n - ans[i] - 1, 0);
  }
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```