#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18, MOD = 1e9 + 7, MOD2 = 1e6 + 3;
const int MAXN = 1e8;
int n, m, tin[1000003], ans[1000003];
int timer, s[1000003], cur;
vector<int> g[1000003];
vector<int> q[1000003];
struct SegTree {
  int t[1000003], ta[1000003], start;
  void build(int n) {
    start = 1;
    while (start < n) start <<= 1;
  }
  void update(int node, int tl, int tr, int l, int r, int d) {
    if (r < tl || tr < l) return;
    if (l <= tl && tr <= r) {
      ta[node] += d;
      if (ta[node])
        t[node] = tr - tl + 1;
      else
        t[node] = t[2 * node] + t[2 * node + 1];
      return;
    }
    int tm = (tl + tr) / 2;
    update(2 * node, tl, tm, l, r, d);
    update(2 * node + 1, tm + 1, tr, l, r, d);
    if (ta[node])
      t[node] = tr - tl + 1;
    else
      t[node] = t[2 * node] + t[2 * node + 1];
  }
  void update(int l, int r, int d) { update(1, 0, start - 1, l, r, d); }
  int get() { return t[1]; }
} t;
int topsort(int x, int p) {
  s[x] = 1;
  tin[x] = timer++;
  for (int to : g[x])
    if (to != p) s[x] += topsort(to, x);
  return s[x];
}
void dfs(int x, int p) {
  if (!q[x].empty()) q[x].push_back(x);
  for (auto a : q[x]) {
    t.update(tin[a], tin[a] + s[a] - 1, 1);
  }
  for (int to : g[x])
    if (to != p) dfs(to, x);
  ans[x] = t.get();
  for (auto a : q[x]) {
    t.update(tin[a], tin[a] + s[a] - 1, -1);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  topsort(0, -1);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    q[a].push_back(b);
    q[b].push_back(a);
  }
  t.build(timer + 1);
  dfs(0, -1);
  for (int i = 0; i < n; i++) {
    cout << max(0, ans[i] - 1) << ' ';
  }
}

### Prompt

Generate 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 long long INF = 1e18, MOD = 1e9 + 7, MOD2 = 1e6 + 3;
const int MAXN = 1e8;
int n, m, tin[1000003], ans[1000003];
int timer, s[1000003], cur;
vector<int> g[1000003];
vector<int> q[1000003];
struct SegTree {
  int t[1000003], ta[1000003], start;
  void build(int n) {
    start = 1;
    while (start < n) start <<= 1;
  }
  void update(int node, int tl, int tr, int l, int r, int d) {
    if (r < tl || tr < l) return;
    if (l <= tl && tr <= r) {
      ta[node] += d;
      if (ta[node])
        t[node] = tr - tl + 1;
      else
        t[node] = t[2 * node] + t[2 * node + 1];
      return;
    }
    int tm = (tl + tr) / 2;
    update(2 * node, tl, tm, l, r, d);
    update(2 * node + 1, tm + 1, tr, l, r, d);
    if (ta[node])
      t[node] = tr - tl + 1;
    else
      t[node] = t[2 * node] + t[2 * node + 1];
  }
  void update(int l, int r, int d) { update(1, 0, start - 1, l, r, d); }
  int get() { return t[1]; }
} t;
int topsort(int x, int p) {
  s[x] = 1;
  tin[x] = timer++;
  for (int to : g[x])
    if (to != p) s[x] += topsort(to, x);
  return s[x];
}
void dfs(int x, int p) {
  if (!q[x].empty()) q[x].push_back(x);
  for (auto a : q[x]) {
    t.update(tin[a], tin[a] + s[a] - 1, 1);
  }
  for (int to : g[x])
    if (to != p) dfs(to, x);
  ans[x] = t.get();
  for (auto a : q[x]) {
    t.update(tin[a], tin[a] + s[a] - 1, -1);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  topsort(0, -1);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    q[a].push_back(b);
    q[b].push_back(a);
  }
  t.build(timer + 1);
  dfs(0, -1);
  for (int i = 0; i < n; i++) {
    cout << max(0, ans[i] - 1) << ' ';
  }
}

```