#include <bits/stdc++.h>
using namespace std;
const int N = 123456;
int cnt;
int tn, tin[N], tout[N], answer[N], cover[N << 2], seg[N << 2];
vector<int> adj[N];
vector<pair<int, int>> queries[N];
void update(int id, int f, int t, int qf, int qt, int v) {
  if (qf <= f && t <= qt) {
    seg[id] += v;
  } else {
    int m = (f + t) / 2;
    if (qf < m + 1) update(id << 1, f, m, qf, qt, v);
    if (qt > m) update(id << 1 | 1, m + 1, t, qf, qt, v);
  }
  if (seg[id] > 0) {
    cover[id] = t - f + 1;
  } else if (f < t) {
    cover[id] = cover[id << 1] + cover[id << 1 | 1];
  } else {
    cover[id] = 0;
  }
}
void dfs(int u, int p) {
  tin[u] = ++tn;
  for (int v : adj[u])
    if (v != p) dfs(v, u);
  tout[u] = tn;
}
void add(int f, int t) { update(1, 1, tn, f, t, 1); }
void rem(int f, int t) { update(1, 1, tn, f, t, -1); }
void dfs_solve(int u, int p) {
  for (auto q : queries[u]) {
    add(tin[q.first], tout[q.first]);
    add(tin[q.second], tout[q.second]);
  }
  if (cover[1] > 0) {
    answer[u] = cover[1] - 1;
  }
  for (int v : adj[u])
    if (v != p) dfs_solve(v, u);
  for (auto q : queries[u]) {
    rem(tin[q.first], tout[q.first]);
    rem(tin[q.second], tout[q.second]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    queries[u].push_back(make_pair(u, v));
    queries[v].push_back(make_pair(u, v));
  }
  dfs_solve(1, 0);
  for (int i = 1; i <= n; ++i) {
    cout << answer[i] << ' ';
  }
}

### 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 = 123456;
int cnt;
int tn, tin[N], tout[N], answer[N], cover[N << 2], seg[N << 2];
vector<int> adj[N];
vector<pair<int, int>> queries[N];
void update(int id, int f, int t, int qf, int qt, int v) {
  if (qf <= f && t <= qt) {
    seg[id] += v;
  } else {
    int m = (f + t) / 2;
    if (qf < m + 1) update(id << 1, f, m, qf, qt, v);
    if (qt > m) update(id << 1 | 1, m + 1, t, qf, qt, v);
  }
  if (seg[id] > 0) {
    cover[id] = t - f + 1;
  } else if (f < t) {
    cover[id] = cover[id << 1] + cover[id << 1 | 1];
  } else {
    cover[id] = 0;
  }
}
void dfs(int u, int p) {
  tin[u] = ++tn;
  for (int v : adj[u])
    if (v != p) dfs(v, u);
  tout[u] = tn;
}
void add(int f, int t) { update(1, 1, tn, f, t, 1); }
void rem(int f, int t) { update(1, 1, tn, f, t, -1); }
void dfs_solve(int u, int p) {
  for (auto q : queries[u]) {
    add(tin[q.first], tout[q.first]);
    add(tin[q.second], tout[q.second]);
  }
  if (cover[1] > 0) {
    answer[u] = cover[1] - 1;
  }
  for (int v : adj[u])
    if (v != p) dfs_solve(v, u);
  for (auto q : queries[u]) {
    rem(tin[q.first], tout[q.first]);
    rem(tin[q.second], tout[q.second]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    queries[u].push_back(make_pair(u, v));
    queries[v].push_back(make_pair(u, v));
  }
  dfs_solve(1, 0);
  for (int i = 1; i <= n; ++i) {
    cout << answer[i] << ' ';
  }
}

```