#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int seg[4 * maxn], lazy[4 * maxn], cnt[4 * maxn];
void propagate(int, int, int);
void change(int id, int L, int R, int l, int r, int val) {
  if (L == l and R == r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  propagate(id, L, R);
  int mid = (L + R) >> 1;
  if (mid > l) change(2 * id + 0, L, mid, l, min(mid, r), val);
  if (mid < r) change(2 * id + 1, mid, R, max(l, mid), r, val);
  seg[id] = min(seg[2 * id + 0], seg[2 * id + 1]);
  cnt[id] = (seg[2 * id + 0] == seg[id]) * cnt[2 * id + 0] +
            (seg[2 * id + 1] == seg[id]) * cnt[2 * id + 1];
}
void propagate(int id, int L, int R) {
  if (lazy[id] == 0) return;
  int mid = (L + R) >> 1;
  change(2 * id + 0, L, mid, L, mid, lazy[id]);
  change(2 * id + 1, mid, R, mid, R, lazy[id]);
  lazy[id] = 0;
}
void build(int id, int L, int R) {
  cnt[id] = R - L;
  if (L + 1 == R) return;
  int mid = (L + R) >> 1;
  build(2 * id + 0, L, mid);
  build(2 * id + 1, mid, R);
}
int Time = 1, st[maxn], ft[maxn];
;
int n;
vector<int> t[maxn];
void dfs_st(int v, int par = -1) {
  st[v] = Time++;
  for (auto u : t[v])
    if (u != par) dfs_st(u, v);
  ft[v] = Time;
}
int answer[maxn];
vector<int> tilnow, List[maxn];
pair<long long, int> q[maxn];
void dfs(int v, int par = -1) {
  for (auto it : List[v]) {
    int u = q[it].first;
    change(1, 1, n + 1, st[u], ft[u], +1);
    u = q[it].second;
    change(1, 1, n + 1, st[u], ft[u], +1);
  }
  if (seg[1] > 0)
    answer[v] = n - 1;
  else
    answer[v] = n - cnt[1] - 1;
  for (auto u : t[v])
    if (u != par) dfs(u, v);
  for (auto it : List[v]) {
    int u = q[it].first;
    change(1, 1, n + 1, st[u], ft[u], -1);
    u = q[it].second;
    change(1, 1, n + 1, st[u], ft[u], -1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    t[v].push_back(u);
    t[u].push_back(v);
  }
  dfs_st(1);
  for (int i = 0; i < m; i++) {
    cin >> q[i].first >> q[i].second;
    List[q[i].first].push_back(i);
    List[q[i].second].push_back(i);
  }
  build(1, 1, n + 1);
  dfs(1);
  for (int i = 1; i <= n; i++) cout << max(answer[i], 0) << " ";
  cout << endl;
}

### Prompt

Please create a solution in Cpp 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 = 1e5 + 10;
int seg[4 * maxn], lazy[4 * maxn], cnt[4 * maxn];
void propagate(int, int, int);
void change(int id, int L, int R, int l, int r, int val) {
  if (L == l and R == r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  propagate(id, L, R);
  int mid = (L + R) >> 1;
  if (mid > l) change(2 * id + 0, L, mid, l, min(mid, r), val);
  if (mid < r) change(2 * id + 1, mid, R, max(l, mid), r, val);
  seg[id] = min(seg[2 * id + 0], seg[2 * id + 1]);
  cnt[id] = (seg[2 * id + 0] == seg[id]) * cnt[2 * id + 0] +
            (seg[2 * id + 1] == seg[id]) * cnt[2 * id + 1];
}
void propagate(int id, int L, int R) {
  if (lazy[id] == 0) return;
  int mid = (L + R) >> 1;
  change(2 * id + 0, L, mid, L, mid, lazy[id]);
  change(2 * id + 1, mid, R, mid, R, lazy[id]);
  lazy[id] = 0;
}
void build(int id, int L, int R) {
  cnt[id] = R - L;
  if (L + 1 == R) return;
  int mid = (L + R) >> 1;
  build(2 * id + 0, L, mid);
  build(2 * id + 1, mid, R);
}
int Time = 1, st[maxn], ft[maxn];
;
int n;
vector<int> t[maxn];
void dfs_st(int v, int par = -1) {
  st[v] = Time++;
  for (auto u : t[v])
    if (u != par) dfs_st(u, v);
  ft[v] = Time;
}
int answer[maxn];
vector<int> tilnow, List[maxn];
pair<long long, int> q[maxn];
void dfs(int v, int par = -1) {
  for (auto it : List[v]) {
    int u = q[it].first;
    change(1, 1, n + 1, st[u], ft[u], +1);
    u = q[it].second;
    change(1, 1, n + 1, st[u], ft[u], +1);
  }
  if (seg[1] > 0)
    answer[v] = n - 1;
  else
    answer[v] = n - cnt[1] - 1;
  for (auto u : t[v])
    if (u != par) dfs(u, v);
  for (auto it : List[v]) {
    int u = q[it].first;
    change(1, 1, n + 1, st[u], ft[u], -1);
    u = q[it].second;
    change(1, 1, n + 1, st[u], ft[u], -1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    t[v].push_back(u);
    t[u].push_back(v);
  }
  dfs_st(1);
  for (int i = 0; i < m; i++) {
    cin >> q[i].first >> q[i].second;
    List[q[i].first].push_back(i);
    List[q[i].second].push_back(i);
  }
  build(1, 1, n + 1);
  dfs(1);
  for (int i = 1; i <= n; i++) cout << max(answer[i], 0) << " ";
  cout << endl;
}

```