#include <bits/stdc++.h>
using namespace std;
long long pw(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y % 2) ret = (ret * x) % 1000000007;
    x = (x * x) % 1000000007;
    y /= 2;
  }
  return ret;
}
long long s[100005] = {0}, e[100005] = {0}, cur = 0;
vector<pair<long long, long long> > rng[100005];
vector<long long> v[100005];
void dfs(long long r, long long p) {
  s[r] = ++cur;
  for (long long i : v[r]) {
    if (i == p) continue;
    dfs(i, r);
  }
  e[r] = cur;
}
long long SGT[4 * 100005] = {0}, LZY[4 * 100005] = {0}, ans[100005] = {0};
void upd(long long id, long long s, long long e, long long l, long long r,
         long long val) {
  if (s > r || l > e) return;
  if (l <= s && e <= r) {
    LZY[id] += val;
    if (LZY[id] > 0)
      SGT[id] = (e - s + 1);
    else {
      if (s != e)
        SGT[id] = SGT[id + id] + SGT[id + id + 1];
      else {
        SGT[id] = 0;
      }
    }
    return;
  }
  long long mid = (s + e) / 2;
  upd(id + id, s, mid, l, r, val);
  upd(id + id + 1, mid + 1, e, l, r, val);
  if (LZY[id])
    SGT[id] = (e - s + 1);
  else
    SGT[id] = SGT[id + id] + SGT[id + id + 1];
}
void dfs1(long long r, long long p) {
  for (auto it : rng[r]) {
    upd(1, 1, 100005 - 1, it.first, it.second, 1);
  }
  ans[r] = SGT[1];
  for (long long i : v[r]) {
    if (i == p) continue;
    dfs1(i, r);
  }
  for (auto it : rng[r]) {
    upd(1, 1, 100005 - 1, it.first, it.second, -1);
  }
}
int main() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  for (long long i = 1; i <= m; i++) {
    long long a, b;
    cin >> a >> b;
    rng[a].push_back({s[a], e[a]});
    rng[a].push_back({s[b], e[b]});
    rng[b].push_back({s[a], e[a]});
    rng[b].push_back({s[b], e[b]});
  }
  dfs1(1, 0);
  for (long long i = 1; i <= n; i++) {
    cout << max(ans[i] - 1, 0ll) << " ";
  }
  cout << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
long long pw(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y % 2) ret = (ret * x) % 1000000007;
    x = (x * x) % 1000000007;
    y /= 2;
  }
  return ret;
}
long long s[100005] = {0}, e[100005] = {0}, cur = 0;
vector<pair<long long, long long> > rng[100005];
vector<long long> v[100005];
void dfs(long long r, long long p) {
  s[r] = ++cur;
  for (long long i : v[r]) {
    if (i == p) continue;
    dfs(i, r);
  }
  e[r] = cur;
}
long long SGT[4 * 100005] = {0}, LZY[4 * 100005] = {0}, ans[100005] = {0};
void upd(long long id, long long s, long long e, long long l, long long r,
         long long val) {
  if (s > r || l > e) return;
  if (l <= s && e <= r) {
    LZY[id] += val;
    if (LZY[id] > 0)
      SGT[id] = (e - s + 1);
    else {
      if (s != e)
        SGT[id] = SGT[id + id] + SGT[id + id + 1];
      else {
        SGT[id] = 0;
      }
    }
    return;
  }
  long long mid = (s + e) / 2;
  upd(id + id, s, mid, l, r, val);
  upd(id + id + 1, mid + 1, e, l, r, val);
  if (LZY[id])
    SGT[id] = (e - s + 1);
  else
    SGT[id] = SGT[id + id] + SGT[id + id + 1];
}
void dfs1(long long r, long long p) {
  for (auto it : rng[r]) {
    upd(1, 1, 100005 - 1, it.first, it.second, 1);
  }
  ans[r] = SGT[1];
  for (long long i : v[r]) {
    if (i == p) continue;
    dfs1(i, r);
  }
  for (auto it : rng[r]) {
    upd(1, 1, 100005 - 1, it.first, it.second, -1);
  }
}
int main() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  for (long long i = 1; i <= m; i++) {
    long long a, b;
    cin >> a >> b;
    rng[a].push_back({s[a], e[a]});
    rng[a].push_back({s[b], e[b]});
    rng[b].push_back({s[a], e[a]});
    rng[b].push_back({s[b], e[b]});
  }
  dfs1(1, 0);
  for (long long i = 1; i <= n; i++) {
    cout << max(ans[i] - 1, 0ll) << " ";
  }
  cout << endl;
  return 0;
}

```