#include <bits/stdc++.h>
using namespace std;
int n, m, c[100100], p[100100], cnt, ans[100100];
int lz[100100 << 2], cc[100100 << 2];
vector<int> g[100100];
vector<int> q[100100];
void dfs(int now, int fa) {
  p[now] = ++cnt;
  c[now] = 1;
  int u;
  for (int i = 0; i < g[now].size(); i++)
    if ((u = g[now][i]) != fa) {
      dfs(u, now);
      c[now] += c[u];
    }
}
void Up(int L, int R, int rt) {
  if (lz[rt])
    cc[rt] = R - L + 1;
  else {
    if (L == R)
      cc[rt] = 0;
    else
      cc[rt] = cc[rt << 1] + cc[rt << 1 | 1];
  }
}
void update(int l, int r, int add, int L, int R, int rt) {
  if (l <= L && R <= r) {
    lz[rt] += add;
    Up(L, R, rt);
    return;
  }
  int mid = (L + R) >> 1;
  if (l <= mid) update(l, r, add, L, mid, rt << 1);
  if (mid < r) update(l, r, add, mid + 1, R, rt << 1 | 1);
  Up(L, R, rt);
}
void work(int now, int fa) {
  int u;
  for (int i = 0; i < q[now].size(); i++) {
    u = q[now][i];
    update(p[u], p[u] + c[u] - 1, 1, 1, n, 1);
  }
  ans[now] = cc[1];
  if (ans[now]) ans[now]--;
  for (int i = 0; i < g[now].size(); i++)
    if ((u = g[now][i]) != fa) {
      work(u, now);
    }
  for (int i = 0; i < q[now].size(); i++) {
    u = q[now][i];
    update(p[u], p[u] + c[u] - 1, -1, 1, n, 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  int x, y;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  cnt = 0;
  dfs(1, 0);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    q[x].push_back(y), q[x].push_back(x);
    q[y].push_back(x), q[y].push_back(y);
  }
  memset(lz, 0, sizeof(lz));
  memset(cc, 0, sizeof(cc));
  work(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  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;
int n, m, c[100100], p[100100], cnt, ans[100100];
int lz[100100 << 2], cc[100100 << 2];
vector<int> g[100100];
vector<int> q[100100];
void dfs(int now, int fa) {
  p[now] = ++cnt;
  c[now] = 1;
  int u;
  for (int i = 0; i < g[now].size(); i++)
    if ((u = g[now][i]) != fa) {
      dfs(u, now);
      c[now] += c[u];
    }
}
void Up(int L, int R, int rt) {
  if (lz[rt])
    cc[rt] = R - L + 1;
  else {
    if (L == R)
      cc[rt] = 0;
    else
      cc[rt] = cc[rt << 1] + cc[rt << 1 | 1];
  }
}
void update(int l, int r, int add, int L, int R, int rt) {
  if (l <= L && R <= r) {
    lz[rt] += add;
    Up(L, R, rt);
    return;
  }
  int mid = (L + R) >> 1;
  if (l <= mid) update(l, r, add, L, mid, rt << 1);
  if (mid < r) update(l, r, add, mid + 1, R, rt << 1 | 1);
  Up(L, R, rt);
}
void work(int now, int fa) {
  int u;
  for (int i = 0; i < q[now].size(); i++) {
    u = q[now][i];
    update(p[u], p[u] + c[u] - 1, 1, 1, n, 1);
  }
  ans[now] = cc[1];
  if (ans[now]) ans[now]--;
  for (int i = 0; i < g[now].size(); i++)
    if ((u = g[now][i]) != fa) {
      work(u, now);
    }
  for (int i = 0; i < q[now].size(); i++) {
    u = q[now][i];
    update(p[u], p[u] + c[u] - 1, -1, 1, n, 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  int x, y;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  cnt = 0;
  dfs(1, 0);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    q[x].push_back(y), q[x].push_back(x);
    q[y].push_back(x), q[y].push_back(y);
  }
  memset(lz, 0, sizeof(lz));
  memset(cc, 0, sizeof(cc));
  work(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

```