#include <bits/stdc++.h>
using namespace std;
const int maxn = 400005;
int v[maxn], l[maxn], r[maxn], ans[maxn];
int tot, n, m, pre[maxn], last[maxn], child[maxn], c[maxn * 2], d[maxn * 2];
vector<int> a[maxn];
void DFSA(int x) {
  if (v[x]) return;
  v[x] = 1;
  l[x] = ++tot;
  for (int i = last[x]; i; i = pre[i]) DFSA(child[i]);
  r[x] = tot;
}
void updata(int i, int len) {
  if (d[i])
    c[i] = len;
  else
    c[i] = c[i * 2] + c[i * 2 + 1];
}
void updata(int i, int a, int b, int l, int r, int delta) {
  if (l <= a && b <= r) {
    d[i] += delta;
    updata(i, b - a + 1);
    return;
  }
  int mid = (a + b) / 2;
  if (l <= mid) updata(i * 2, a, mid, l, r, delta);
  if (mid < r) updata(i * 2 + 1, mid + 1, b, l, r, delta);
  updata(i, b - a + 1);
}
void DFSB(int x) {
  if (v[x]) return;
  v[x] = 1;
  for (int i = 0; i < a[x].size(); i++)
    updata(1, 1, n, l[a[x][i]], r[a[x][i]], 1);
  ans[x] = max(c[1] - 1, 0);
  for (int i = last[x]; i; i = pre[i]) DFSB(child[i]);
  for (int i = 0; i < a[x].size(); i++)
    updata(1, 1, n, l[a[x][i]], r[a[x][i]], -1);
}
int main() {
  cin >> n >> m;
  for (int i = 1, t = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    pre[++t] = last[x];
    last[x] = t;
    child[t] = y;
    pre[++t] = last[y];
    last[y] = t;
    child[t] = x;
  }
  DFSA(1);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x].push_back(x);
    a[x].push_back(y);
    a[y].push_back(x);
    a[y].push_back(y);
  }
  memset(v, 0, sizeof(v));
  DFSB(1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 400005;
int v[maxn], l[maxn], r[maxn], ans[maxn];
int tot, n, m, pre[maxn], last[maxn], child[maxn], c[maxn * 2], d[maxn * 2];
vector<int> a[maxn];
void DFSA(int x) {
  if (v[x]) return;
  v[x] = 1;
  l[x] = ++tot;
  for (int i = last[x]; i; i = pre[i]) DFSA(child[i]);
  r[x] = tot;
}
void updata(int i, int len) {
  if (d[i])
    c[i] = len;
  else
    c[i] = c[i * 2] + c[i * 2 + 1];
}
void updata(int i, int a, int b, int l, int r, int delta) {
  if (l <= a && b <= r) {
    d[i] += delta;
    updata(i, b - a + 1);
    return;
  }
  int mid = (a + b) / 2;
  if (l <= mid) updata(i * 2, a, mid, l, r, delta);
  if (mid < r) updata(i * 2 + 1, mid + 1, b, l, r, delta);
  updata(i, b - a + 1);
}
void DFSB(int x) {
  if (v[x]) return;
  v[x] = 1;
  for (int i = 0; i < a[x].size(); i++)
    updata(1, 1, n, l[a[x][i]], r[a[x][i]], 1);
  ans[x] = max(c[1] - 1, 0);
  for (int i = last[x]; i; i = pre[i]) DFSB(child[i]);
  for (int i = 0; i < a[x].size(); i++)
    updata(1, 1, n, l[a[x][i]], r[a[x][i]], -1);
}
int main() {
  cin >> n >> m;
  for (int i = 1, t = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    pre[++t] = last[x];
    last[x] = t;
    child[t] = y;
    pre[++t] = last[y];
    last[y] = t;
    child[t] = x;
  }
  DFSA(1);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x].push_back(x);
    a[x].push_back(y);
    a[y].push_back(x);
    a[y].push_back(y);
  }
  memset(v, 0, sizeof(v));
  DFSB(1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}

```