#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, pos[N], size[N], lazy[4 * N], now = 0, ans[N];
pair<int, int> it[4 * N];
vector<int> que[N], adj[N];
void dfs(int x, int p) {
  now++;
  pos[x] = now;
  size[x] = 1;
  for (int i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p) {
      dfs(adj[x][i], x);
      size[x] += size[adj[x][i]];
    }
  }
}
void init(int idx, int l, int r) {
  if (l == r) {
    it[idx].first = 0;
    it[idx].second = 1;
    return;
  }
  it[idx].first = 0;
  it[idx].second = r - l + 1;
  init(2 * idx, l, (l + r) / 2);
  init(2 * idx + 1, (l + r) / 2 + 1, r);
}
void upd(int idx, int l, int r, int lef, int rig, int val) {
  if (l > rig || r < lef) {
    return;
  }
  if (lef > rig) {
    return;
  }
  if (l >= lef && r <= rig) {
    it[idx].first += val;
    lazy[idx] += val;
    return;
  }
  if (lazy[idx] != 0) {
    it[2 * idx].first += lazy[idx];
    lazy[2 * idx] += lazy[idx];
    it[2 * idx + 1].first += lazy[idx];
    lazy[2 * idx + 1] += lazy[idx];
    lazy[idx] = 0;
  }
  upd(2 * idx, l, (l + r) / 2, lef, rig, val);
  upd(2 * idx + 1, (l + r) / 2 + 1, r, lef, rig, val);
  if (it[2 * idx].first > it[2 * idx + 1].first) {
    it[idx].first = it[2 * idx + 1].first;
    it[idx].second = it[2 * idx + 1].second;
  }
  if (it[2 * idx].first == it[2 * idx + 1].first) {
    it[idx].first = it[2 * idx + 1].first;
    it[idx].second = it[2 * idx].second + it[2 * idx + 1].second;
  }
  if (it[2 * idx].first < it[2 * idx + 1].first) {
    it[idx].first = it[2 * idx].first;
    it[idx].second = it[2 * idx].second;
  }
}
int getsum(int idx, int l, int r, int pos1) {
  if (l > pos1 || r < pos1) {
    return 0;
  }
  if (l == r) {
    return it[idx].first;
  }
  if (lazy[idx] != 0) {
    it[2 * idx].first += lazy[idx];
    lazy[2 * idx] += lazy[idx];
    it[2 * idx + 1].first += lazy[idx];
    lazy[2 * idx + 1] += lazy[idx];
    lazy[idx] = 0;
  }
  return getsum(2 * idx, l, (l + r) / 2, pos1) +
         getsum(2 * idx + 1, (l + r) / 2 + 1, r, pos1);
}
void getans(int x, int p) {
  int i;
  if (que[x].size()) {
    upd(1, 1, n, pos[x], pos[x] + size[x] - 1, 1);
  }
  for (i = 0; i < que[x].size(); i++) {
    upd(1, 1, n, pos[que[x][i]], pos[que[x][i]] + size[que[x][i]] - 1, 1);
  }
  if (it[1].first == 0) {
    ans[x] = n - it[1].second;
  } else {
    ans[x] = n;
  }
  if (getsum(1, 1, n, pos[x])) {
    ans[x]--;
  }
  for (i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p) {
      getans(adj[x][i], x);
    }
  }
  if (que[x].size()) {
    upd(1, 1, n, pos[x], pos[x] + size[x] - 1, -1);
  }
  for (i = 0; i < que[x].size(); i++) {
    upd(1, 1, n, pos[que[x][i]], pos[que[x][i]] + size[que[x][i]] - 1, -1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int i, j, k, l, q;
  cin >> n >> q;
  for (i = 1; i < n; i++) {
    cin >> j >> k;
    adj[j].push_back(k);
    adj[k].push_back(j);
  }
  for (i = 1; i <= q; i++) {
    cin >> j >> k;
    que[j].push_back(k);
    que[k].push_back(j);
  }
  dfs(1, 1);
  init(1, 1, n);
  getans(1, 1);
  for (i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
}

### 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 N = 1e5 + 5;
int n, pos[N], size[N], lazy[4 * N], now = 0, ans[N];
pair<int, int> it[4 * N];
vector<int> que[N], adj[N];
void dfs(int x, int p) {
  now++;
  pos[x] = now;
  size[x] = 1;
  for (int i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p) {
      dfs(adj[x][i], x);
      size[x] += size[adj[x][i]];
    }
  }
}
void init(int idx, int l, int r) {
  if (l == r) {
    it[idx].first = 0;
    it[idx].second = 1;
    return;
  }
  it[idx].first = 0;
  it[idx].second = r - l + 1;
  init(2 * idx, l, (l + r) / 2);
  init(2 * idx + 1, (l + r) / 2 + 1, r);
}
void upd(int idx, int l, int r, int lef, int rig, int val) {
  if (l > rig || r < lef) {
    return;
  }
  if (lef > rig) {
    return;
  }
  if (l >= lef && r <= rig) {
    it[idx].first += val;
    lazy[idx] += val;
    return;
  }
  if (lazy[idx] != 0) {
    it[2 * idx].first += lazy[idx];
    lazy[2 * idx] += lazy[idx];
    it[2 * idx + 1].first += lazy[idx];
    lazy[2 * idx + 1] += lazy[idx];
    lazy[idx] = 0;
  }
  upd(2 * idx, l, (l + r) / 2, lef, rig, val);
  upd(2 * idx + 1, (l + r) / 2 + 1, r, lef, rig, val);
  if (it[2 * idx].first > it[2 * idx + 1].first) {
    it[idx].first = it[2 * idx + 1].first;
    it[idx].second = it[2 * idx + 1].second;
  }
  if (it[2 * idx].first == it[2 * idx + 1].first) {
    it[idx].first = it[2 * idx + 1].first;
    it[idx].second = it[2 * idx].second + it[2 * idx + 1].second;
  }
  if (it[2 * idx].first < it[2 * idx + 1].first) {
    it[idx].first = it[2 * idx].first;
    it[idx].second = it[2 * idx].second;
  }
}
int getsum(int idx, int l, int r, int pos1) {
  if (l > pos1 || r < pos1) {
    return 0;
  }
  if (l == r) {
    return it[idx].first;
  }
  if (lazy[idx] != 0) {
    it[2 * idx].first += lazy[idx];
    lazy[2 * idx] += lazy[idx];
    it[2 * idx + 1].first += lazy[idx];
    lazy[2 * idx + 1] += lazy[idx];
    lazy[idx] = 0;
  }
  return getsum(2 * idx, l, (l + r) / 2, pos1) +
         getsum(2 * idx + 1, (l + r) / 2 + 1, r, pos1);
}
void getans(int x, int p) {
  int i;
  if (que[x].size()) {
    upd(1, 1, n, pos[x], pos[x] + size[x] - 1, 1);
  }
  for (i = 0; i < que[x].size(); i++) {
    upd(1, 1, n, pos[que[x][i]], pos[que[x][i]] + size[que[x][i]] - 1, 1);
  }
  if (it[1].first == 0) {
    ans[x] = n - it[1].second;
  } else {
    ans[x] = n;
  }
  if (getsum(1, 1, n, pos[x])) {
    ans[x]--;
  }
  for (i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p) {
      getans(adj[x][i], x);
    }
  }
  if (que[x].size()) {
    upd(1, 1, n, pos[x], pos[x] + size[x] - 1, -1);
  }
  for (i = 0; i < que[x].size(); i++) {
    upd(1, 1, n, pos[que[x][i]], pos[que[x][i]] + size[que[x][i]] - 1, -1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int i, j, k, l, q;
  cin >> n >> q;
  for (i = 1; i < n; i++) {
    cin >> j >> k;
    adj[j].push_back(k);
    adj[k].push_back(j);
  }
  for (i = 1; i <= q; i++) {
    cin >> j >> k;
    que[j].push_back(k);
    que[k].push_back(j);
  }
  dfs(1, 1);
  init(1, 1, n);
  getans(1, 1);
  for (i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
}

```