#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000 + 10;
const int MAXT = (1 << 18);
bool mark[MAXN];
int n, que, now;
int c[MAXN], lo[MAXN], hi[MAXN];
int flag[MAXT], dp[MAXT];
vector<int> Q[MAXN], adj[MAXN];
inline int dfs(int v) {
  mark[v] = true;
  lo[v] = hi[v] = now++;
  for (int i = 0; i < (int)adj[v].size(); i++)
    if (!mark[adj[v][i]]) hi[v] = dfs(adj[v][i]);
  return hi[v];
}
inline void add(int root, int lo, int hi, int l, int r, int cost) {
  if (r < lo || hi < l) return;
  if (l <= lo && hi <= r) {
    flag[root] += cost;
    if (flag[root] > 0)
      dp[root] = hi - lo + 1;
    else
      dp[root] = (lo != hi) ? (dp[root << 1] + dp[(root << 1) + 1]) : 0;
    return;
  }
  int mid = (lo + hi) / 2;
  add(root << 1, lo, mid, l, r, cost);
  add((root << 1) + 1, mid + 1, hi, l, r, cost);
  dp[root] =
      (flag[root] == 0) ? (dp[root << 1] + dp[(root << 1) + 1]) : (hi - lo + 1);
}
inline void go(int v) {
  mark[v] = true;
  for (int i = 0; i < (int)Q[v].size(); i++)
    add(1, 0, n - 1, lo[Q[v][i]], hi[Q[v][i]], 1);
  if (!Q[v].empty()) add(1, 0, n - 1, lo[v], hi[v], 1);
  c[v] = max(0, dp[1] - 1);
  for (int i = 0; i < (int)adj[v].size(); i++)
    if (!mark[adj[v][i]]) go(adj[v][i]);
  for (int i = 0; i < (int)Q[v].size(); i++)
    add(1, 0, n - 1, lo[Q[v][i]], hi[Q[v][i]], -1);
  if (!Q[v].empty()) add(1, 0, n - 1, lo[v], hi[v], -1);
}
int main() {
  cin >> n >> que;
  for (int i = 0; i < n - 1; i++) {
    int e1, e2;
    cin >> e1 >> e2;
    e1--;
    e2--;
    adj[e1].push_back(e2);
    adj[e2].push_back(e1);
  }
  dfs(0);
  for (int o = 0; o < que; o++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    Q[a].push_back(b);
    Q[b].push_back(a);
  }
  memset(mark, false, sizeof mark);
  go(0);
  for (int i = 0; i < n; i++) cout << c[i] << ' ';
  cout << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 100000 + 10;
const int MAXT = (1 << 18);
bool mark[MAXN];
int n, que, now;
int c[MAXN], lo[MAXN], hi[MAXN];
int flag[MAXT], dp[MAXT];
vector<int> Q[MAXN], adj[MAXN];
inline int dfs(int v) {
  mark[v] = true;
  lo[v] = hi[v] = now++;
  for (int i = 0; i < (int)adj[v].size(); i++)
    if (!mark[adj[v][i]]) hi[v] = dfs(adj[v][i]);
  return hi[v];
}
inline void add(int root, int lo, int hi, int l, int r, int cost) {
  if (r < lo || hi < l) return;
  if (l <= lo && hi <= r) {
    flag[root] += cost;
    if (flag[root] > 0)
      dp[root] = hi - lo + 1;
    else
      dp[root] = (lo != hi) ? (dp[root << 1] + dp[(root << 1) + 1]) : 0;
    return;
  }
  int mid = (lo + hi) / 2;
  add(root << 1, lo, mid, l, r, cost);
  add((root << 1) + 1, mid + 1, hi, l, r, cost);
  dp[root] =
      (flag[root] == 0) ? (dp[root << 1] + dp[(root << 1) + 1]) : (hi - lo + 1);
}
inline void go(int v) {
  mark[v] = true;
  for (int i = 0; i < (int)Q[v].size(); i++)
    add(1, 0, n - 1, lo[Q[v][i]], hi[Q[v][i]], 1);
  if (!Q[v].empty()) add(1, 0, n - 1, lo[v], hi[v], 1);
  c[v] = max(0, dp[1] - 1);
  for (int i = 0; i < (int)adj[v].size(); i++)
    if (!mark[adj[v][i]]) go(adj[v][i]);
  for (int i = 0; i < (int)Q[v].size(); i++)
    add(1, 0, n - 1, lo[Q[v][i]], hi[Q[v][i]], -1);
  if (!Q[v].empty()) add(1, 0, n - 1, lo[v], hi[v], -1);
}
int main() {
  cin >> n >> que;
  for (int i = 0; i < n - 1; i++) {
    int e1, e2;
    cin >> e1 >> e2;
    e1--;
    e2--;
    adj[e1].push_back(e2);
    adj[e2].push_back(e1);
  }
  dfs(0);
  for (int o = 0; o < que; o++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    Q[a].push_back(b);
    Q[b].push_back(a);
  }
  memset(mark, false, sizeof mark);
  go(0);
  for (int i = 0; i < n; i++) cout << c[i] << ' ';
  cout << endl;
  return 0;
}

```