#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int sum[N << 2], lazy[N << 2], mx[N << 2], dfn[N], ed[N], cnt, head[N], tot, n,
    m, ans[N];
struct edge {
  int next, to;
} e[N << 1];
vector<int> V[N];
void add(int u, int v) {
  e[++cnt].next = head[u];
  e[cnt].to = v;
  head[u] = cnt;
}
void push_down(int rt, int l, int r) {
  if (lazy[rt]) {
    int mid = l + r >> 1;
    sum[(rt << 1)] = mid - l + 1;
    sum[(rt << 1 | 1)] = r - mid;
    lazy[(rt << 1)] += lazy[rt];
    lazy[(rt << 1 | 1)] += lazy[rt];
    mx[(rt << 1)] += lazy[rt];
    mx[(rt << 1 | 1)] += lazy[rt];
    lazy[rt] = 0;
  }
}
void push_up(int rt) {
  sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)];
  mx[rt] = min(mx[(rt << 1)], mx[(rt << 1 | 1)]);
}
void update(int l, int r, int L, int R, int rt) {
  if (L <= l && r <= R) {
    sum[rt] = r - l + 1;
    lazy[rt] += 1;
    mx[rt] += 1;
    return;
  }
  int mid = l + r >> 1;
  push_down(rt, l, r);
  if (L <= mid) update(l, mid, L, R, (rt << 1));
  if (mid < R) update(mid + 1, r, L, R, (rt << 1 | 1));
  push_up(rt);
}
void up1(int l, int r, int L, int R, int rt) {
  if (L <= l && r <= R && mx[rt] > 1) {
    mx[rt] -= 1;
    lazy[rt] -= 1;
    return;
  }
  if (l == r) {
    mx[rt] = 0;
    sum[rt] = 0;
    return;
  }
  push_down(rt, l, r);
  int mid = l + r >> 1;
  if (L <= mid) up1(l, mid, L, R, (rt << 1));
  if (mid < R) up1(mid + 1, r, L, R, (rt << 1 | 1));
  push_up(rt);
}
void dfs(int u, int f) {
  dfn[u] = ++tot;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == f) continue;
    dfs(v, u);
  }
  ed[u] = tot;
}
void dfs1(int u, int f) {
  for (auto it : V[u]) {
    update(1, n, dfn[it], ed[it], 1);
    update(1, n, dfn[u], ed[u], 1);
  }
  ans[u] = max(1, sum[1]);
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == f) continue;
    dfs1(v, u);
  }
  for (auto it : V[u]) {
    up1(1, n, dfn[it], ed[it], 1);
    up1(1, n, dfn[u], ed[u], 1);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    add(a, b);
    add(b, a);
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    V[x].push_back(y);
    V[y].push_back(x);
  }
  dfs(1, 0);
  dfs1(1, 0);
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i] - 1, i == n ? '\n' : ' ');
  return 0;
}

### Prompt

Your challenge is to write 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 N = 1e5 + 5;
int sum[N << 2], lazy[N << 2], mx[N << 2], dfn[N], ed[N], cnt, head[N], tot, n,
    m, ans[N];
struct edge {
  int next, to;
} e[N << 1];
vector<int> V[N];
void add(int u, int v) {
  e[++cnt].next = head[u];
  e[cnt].to = v;
  head[u] = cnt;
}
void push_down(int rt, int l, int r) {
  if (lazy[rt]) {
    int mid = l + r >> 1;
    sum[(rt << 1)] = mid - l + 1;
    sum[(rt << 1 | 1)] = r - mid;
    lazy[(rt << 1)] += lazy[rt];
    lazy[(rt << 1 | 1)] += lazy[rt];
    mx[(rt << 1)] += lazy[rt];
    mx[(rt << 1 | 1)] += lazy[rt];
    lazy[rt] = 0;
  }
}
void push_up(int rt) {
  sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)];
  mx[rt] = min(mx[(rt << 1)], mx[(rt << 1 | 1)]);
}
void update(int l, int r, int L, int R, int rt) {
  if (L <= l && r <= R) {
    sum[rt] = r - l + 1;
    lazy[rt] += 1;
    mx[rt] += 1;
    return;
  }
  int mid = l + r >> 1;
  push_down(rt, l, r);
  if (L <= mid) update(l, mid, L, R, (rt << 1));
  if (mid < R) update(mid + 1, r, L, R, (rt << 1 | 1));
  push_up(rt);
}
void up1(int l, int r, int L, int R, int rt) {
  if (L <= l && r <= R && mx[rt] > 1) {
    mx[rt] -= 1;
    lazy[rt] -= 1;
    return;
  }
  if (l == r) {
    mx[rt] = 0;
    sum[rt] = 0;
    return;
  }
  push_down(rt, l, r);
  int mid = l + r >> 1;
  if (L <= mid) up1(l, mid, L, R, (rt << 1));
  if (mid < R) up1(mid + 1, r, L, R, (rt << 1 | 1));
  push_up(rt);
}
void dfs(int u, int f) {
  dfn[u] = ++tot;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == f) continue;
    dfs(v, u);
  }
  ed[u] = tot;
}
void dfs1(int u, int f) {
  for (auto it : V[u]) {
    update(1, n, dfn[it], ed[it], 1);
    update(1, n, dfn[u], ed[u], 1);
  }
  ans[u] = max(1, sum[1]);
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == f) continue;
    dfs1(v, u);
  }
  for (auto it : V[u]) {
    up1(1, n, dfn[it], ed[it], 1);
    up1(1, n, dfn[u], ed[u], 1);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    add(a, b);
    add(b, a);
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    V[x].push_back(y);
    V[y].push_back(x);
  }
  dfs(1, 0);
  dfs1(1, 0);
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i] - 1, i == n ? '\n' : ' ');
  return 0;
}

```