#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int maxm = 200005;
const int inf = 1e9 + 7;
vector<int> hav[maxn];
int st[maxn], en[maxn], tot, cnt, ans[maxn];
int n, m, head[maxn], to[maxm], nex[maxm];
int sum[maxn << 3], laz[maxn << 3];
void push_up(int l, int r, int rt) {
  if (laz[rt])
    sum[rt] = r - l + 1;
  else {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  }
}
void update(int l, int r, int rt, int ql, int qr, int v) {
  if (ql <= l && r <= qr) {
    laz[rt] += v;
    push_up(l, r, rt);
    return;
  }
  if (ql <= (l + r) / 2) update(l, (l + r) / 2, rt << 1, ql, qr, v);
  if (qr > (l + r) / 2) update((l + r) / 2 + 1, r, rt << 1 | 1, ql, qr, v);
  push_up(l, r, rt);
}
void add(int u, int v) {
  to[cnt] = v;
  nex[cnt] = head[u];
  head[u] = cnt++;
}
void init(int u, int f) {
  st[u] = ++tot;
  for (int i = head[u]; ~i; i = nex[i]) {
    int v = to[i];
    if (v == f) continue;
    init(v, u);
  }
  en[u] = tot;
}
void dfs(int u, int f) {
  for (auto x : hav[u]) {
    update(1, tot, 1, st[x], en[x], 1);
  }
  ans[u] = max(sum[1] - 1, 0);
  for (int i = head[u]; ~i; i = nex[i]) {
    int v = to[i];
    if (f == v) continue;
    dfs(v, u);
  }
  for (auto x : hav[u]) {
    update(1, tot, 1, st[x], en[x], -1);
  }
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &n, &m);
  for (int i = (int)2; i <= (int)n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  for (int i = (int)1; i <= (int)m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    hav[a].push_back(a);
    hav[b].push_back(a);
    hav[a].push_back(b);
    hav[b].push_back(b);
  }
  init(1, -1);
  dfs(1, -1);
  for (int i = (int)1; i <= (int)n; i++) {
    printf("%d%c", ans[i], i == n ? '\n' : ' ');
  }
  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 = 100005;
const int maxm = 200005;
const int inf = 1e9 + 7;
vector<int> hav[maxn];
int st[maxn], en[maxn], tot, cnt, ans[maxn];
int n, m, head[maxn], to[maxm], nex[maxm];
int sum[maxn << 3], laz[maxn << 3];
void push_up(int l, int r, int rt) {
  if (laz[rt])
    sum[rt] = r - l + 1;
  else {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  }
}
void update(int l, int r, int rt, int ql, int qr, int v) {
  if (ql <= l && r <= qr) {
    laz[rt] += v;
    push_up(l, r, rt);
    return;
  }
  if (ql <= (l + r) / 2) update(l, (l + r) / 2, rt << 1, ql, qr, v);
  if (qr > (l + r) / 2) update((l + r) / 2 + 1, r, rt << 1 | 1, ql, qr, v);
  push_up(l, r, rt);
}
void add(int u, int v) {
  to[cnt] = v;
  nex[cnt] = head[u];
  head[u] = cnt++;
}
void init(int u, int f) {
  st[u] = ++tot;
  for (int i = head[u]; ~i; i = nex[i]) {
    int v = to[i];
    if (v == f) continue;
    init(v, u);
  }
  en[u] = tot;
}
void dfs(int u, int f) {
  for (auto x : hav[u]) {
    update(1, tot, 1, st[x], en[x], 1);
  }
  ans[u] = max(sum[1] - 1, 0);
  for (int i = head[u]; ~i; i = nex[i]) {
    int v = to[i];
    if (f == v) continue;
    dfs(v, u);
  }
  for (auto x : hav[u]) {
    update(1, tot, 1, st[x], en[x], -1);
  }
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &n, &m);
  for (int i = (int)2; i <= (int)n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  for (int i = (int)1; i <= (int)m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    hav[a].push_back(a);
    hav[b].push_back(a);
    hav[a].push_back(b);
    hav[b].push_back(b);
  }
  init(1, -1);
  dfs(1, -1);
  for (int i = (int)1; i <= (int)n; i++) {
    printf("%d%c", ans[i], i == n ? '\n' : ' ');
  }
  return 0;
}

```