#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
vector<int> E[MAXN], S[MAXN];
int L[MAXN], R[MAXN], id, ans[MAXN];
int N, M, u, v;
int Tree[MAXN << 2], Sum[MAXN << 2];
void add(int a, int b, int v, int pos = 1, int l = 1, int r = N) {
  Sum[pos] += (b - a + 1) * v;
  if (a == l && b == r) {
    Tree[pos] += v;
    return;
  }
  int Mid = (l + r) >> 1;
  if (b <= Mid) return add(a, b, v, pos * 2, l, Mid);
  if (a > Mid) return add(a, b, v, pos * 2 + 1, Mid + 1, r);
  add(a, Mid, v, pos * 2, l, Mid);
  add(Mid + 1, b, v, pos * 2 + 1, Mid + 1, r);
}
int query(int pos = 1, int l = 1, int r = N) {
  if (!Sum[pos]) return 0;
  if (Tree[pos]) return r - l + 1;
  int Mid = (l + r) >> 1;
  return query(pos * 2, l, Mid) + query(pos * 2 + 1, Mid + 1, r);
}
void dfs1(int u = 1, int pre = 0) {
  L[u] = ++id;
  for (int i = E[u].size() - 1; ~i; --i) {
    if (E[u][i] == pre) continue;
    dfs1(E[u][i], u);
  }
  R[u] = id;
}
void dfs2(int u = 1, int pre = 0) {
  for (int i = S[u].size() - 1; ~i; --i) {
    add(L[S[u][i]], R[S[u][i]], 1);
  }
  add(L[u], L[u], 1);
  ans[u] = query() - 1;
  add(L[u], L[u], -1);
  for (int i = E[u].size() - 1; ~i; --i) {
    if (E[u][i] == pre) continue;
    dfs2(E[u][i], u);
  }
  for (int i = S[u].size() - 1; ~i; --i) add(L[S[u][i]], R[S[u][i]], -1);
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i < N; ++i) {
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  dfs1();
  while (M--) {
    scanf("%d%d", &u, &v);
    S[u].push_back(v);
    S[u].push_back(u);
    S[v].push_back(u);
    S[v].push_back(v);
  }
  dfs2();
  for (int i = 1; i < N; ++i) printf("%d ", ans[i]);
  printf("%d\n", ans[N]);
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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 MAXN = 100005;
vector<int> E[MAXN], S[MAXN];
int L[MAXN], R[MAXN], id, ans[MAXN];
int N, M, u, v;
int Tree[MAXN << 2], Sum[MAXN << 2];
void add(int a, int b, int v, int pos = 1, int l = 1, int r = N) {
  Sum[pos] += (b - a + 1) * v;
  if (a == l && b == r) {
    Tree[pos] += v;
    return;
  }
  int Mid = (l + r) >> 1;
  if (b <= Mid) return add(a, b, v, pos * 2, l, Mid);
  if (a > Mid) return add(a, b, v, pos * 2 + 1, Mid + 1, r);
  add(a, Mid, v, pos * 2, l, Mid);
  add(Mid + 1, b, v, pos * 2 + 1, Mid + 1, r);
}
int query(int pos = 1, int l = 1, int r = N) {
  if (!Sum[pos]) return 0;
  if (Tree[pos]) return r - l + 1;
  int Mid = (l + r) >> 1;
  return query(pos * 2, l, Mid) + query(pos * 2 + 1, Mid + 1, r);
}
void dfs1(int u = 1, int pre = 0) {
  L[u] = ++id;
  for (int i = E[u].size() - 1; ~i; --i) {
    if (E[u][i] == pre) continue;
    dfs1(E[u][i], u);
  }
  R[u] = id;
}
void dfs2(int u = 1, int pre = 0) {
  for (int i = S[u].size() - 1; ~i; --i) {
    add(L[S[u][i]], R[S[u][i]], 1);
  }
  add(L[u], L[u], 1);
  ans[u] = query() - 1;
  add(L[u], L[u], -1);
  for (int i = E[u].size() - 1; ~i; --i) {
    if (E[u][i] == pre) continue;
    dfs2(E[u][i], u);
  }
  for (int i = S[u].size() - 1; ~i; --i) add(L[S[u][i]], R[S[u][i]], -1);
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i < N; ++i) {
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  dfs1();
  while (M--) {
    scanf("%d%d", &u, &v);
    S[u].push_back(v);
    S[u].push_back(u);
    S[v].push_back(u);
    S[v].push_back(v);
  }
  dfs2();
  for (int i = 1; i < N; ++i) printf("%d ", ans[i]);
  printf("%d\n", ans[N]);
  return 0;
}

```