#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
struct Node {
  int l, r, w, s;
} t[N << 2];
vector<int> a[N], c[N];
int p[N], q[N], L[N], R[N], s[N], z[N];
int cnt, n, m, x, y;
void update(int x) {
  if (t[x].w)
    t[x].s = t[x].r - t[x].l + 1;
  else if (t[x].r != t[x].l)
    t[x].s = t[x << 1].s + t[x << 1 | 1].s;
  else
    t[x].s = 0;
}
void build(int x, int l, int r) {
  t[x].l = l, t[x].r = r;
  if (l == r) return;
  int m = (l + r) >> 1;
  build(x << 1, l, m);
  build(x << 1 | 1, m + 1, r);
  update(x);
}
void change(int x, int l, int r, int d) {
  if (t[x].l > r || t[x].r < l)
    return;
  else if (l <= t[x].l && t[x].r <= r)
    t[x].w += d;
  else
    change(x << 1, l, r, d), change(x << 1 | 1, l, r, d);
  update(x);
}
void dfs(int x, int y) {
  L[x] = ++cnt;
  for (__typeof(a[x].begin()) i = a[x].begin(); i != a[x].end(); i++)
    if (*i != y) dfs(*i, x);
  R[x] = cnt;
}
inline void tcg(int k, int d) { change(1, L[k], R[k], d); }
void DFS(int x, int y) {
  for (__typeof(c[x].begin()) i = c[x].begin(); i != c[x].end(); i++)
    tcg(p[*i], 1), tcg(q[*i], 1);
  z[x] = t[1].s;
  for (__typeof(a[x].begin()) i = a[x].begin(); i != a[x].end(); i++)
    if (*i != y) DFS(*i, x);
  for (__typeof(c[x].begin()) i = c[x].begin(); i != c[x].end(); i++)
    tcg(p[*i], -1), tcg(q[*i], -1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  build(1, 1, n);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", p + i, q + i);
    c[p[i]].push_back(i);
    c[q[i]].push_back(i);
  }
  DFS(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", max(z[i] - 1, 0));
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 N = 100010;
struct Node {
  int l, r, w, s;
} t[N << 2];
vector<int> a[N], c[N];
int p[N], q[N], L[N], R[N], s[N], z[N];
int cnt, n, m, x, y;
void update(int x) {
  if (t[x].w)
    t[x].s = t[x].r - t[x].l + 1;
  else if (t[x].r != t[x].l)
    t[x].s = t[x << 1].s + t[x << 1 | 1].s;
  else
    t[x].s = 0;
}
void build(int x, int l, int r) {
  t[x].l = l, t[x].r = r;
  if (l == r) return;
  int m = (l + r) >> 1;
  build(x << 1, l, m);
  build(x << 1 | 1, m + 1, r);
  update(x);
}
void change(int x, int l, int r, int d) {
  if (t[x].l > r || t[x].r < l)
    return;
  else if (l <= t[x].l && t[x].r <= r)
    t[x].w += d;
  else
    change(x << 1, l, r, d), change(x << 1 | 1, l, r, d);
  update(x);
}
void dfs(int x, int y) {
  L[x] = ++cnt;
  for (__typeof(a[x].begin()) i = a[x].begin(); i != a[x].end(); i++)
    if (*i != y) dfs(*i, x);
  R[x] = cnt;
}
inline void tcg(int k, int d) { change(1, L[k], R[k], d); }
void DFS(int x, int y) {
  for (__typeof(c[x].begin()) i = c[x].begin(); i != c[x].end(); i++)
    tcg(p[*i], 1), tcg(q[*i], 1);
  z[x] = t[1].s;
  for (__typeof(a[x].begin()) i = a[x].begin(); i != a[x].end(); i++)
    if (*i != y) DFS(*i, x);
  for (__typeof(c[x].begin()) i = c[x].begin(); i != c[x].end(); i++)
    tcg(p[*i], -1), tcg(q[*i], -1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  build(1, 1, n);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", p + i, q + i);
    c[p[i]].push_back(i);
    c[q[i]].push_back(i);
  }
  DFS(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", max(z[i] - 1, 0));
  return 0;
}

```