#include <bits/stdc++.h>
using namespace std;
int cover[1 << 18], sum[1 << 18], ans[100000], loc[100000], size[100000],
    depth[100000], pre[100000][20];
vector<int> ge[100000], l[100000];
bool v[100000];
void dfs(int i) {
  static int cur = 0;
  loc[i] = cur++;
  size[i] = 1;
  v[i] = true;
  for (int j = 0; j < ge[i].size(); j++) {
    int son = ge[i][j];
    if (v[son]) continue;
    depth[son] = depth[i] + 1;
    pre[son][0] = i;
    for (int k = 1; 1 << k <= depth[son]; k++)
      pre[son][k] = pre[pre[son][k - 1]][k - 1];
    dfs(son);
    size[i] += size[son];
  }
}
int getLca(int a, int b) {
  if (depth[a] < depth[b]) swap(a, b);
  int lg = 0;
  while (1 << lg <= depth[a]) lg++;
  for (int i = lg - 1; i >= 0; i--)
    if (depth[a] - (1 << i) >= depth[b]) a = pre[a][i];
  if (a == b) return a;
  for (int i = lg - 1; i >= 0; i--)
    if (pre[a][i] != -1 && pre[a][i] != pre[b][i]) {
      a = pre[a][i];
      b = pre[b][i];
    }
  return pre[a][0];
}
void update(int i, int l, int r) {
  if (cover[i] == 0) {
    if (l + 1 == r) {
      sum[i] = 0;
    } else {
      sum[i] = sum[i * 2] + sum[i * 2 + 1];
    }
  } else {
    sum[i] = r - l;
  }
}
void coverRange(int i, int l, int r, int l0, int r0, int d) {
  if (l0 <= l && r <= r0) {
    cover[i] += d;
    update(i, l, r);
    return;
  }
  int mid = (l + r) / 2;
  if (l0 < mid) coverRange(i * 2, l, mid, l0, r0, d);
  if (r0 > mid) coverRange(i * 2 + 1, mid, r, l0, r0, d);
  update(i, l, r);
}
void solve(int i, int n) {
  v[i] = true;
  for (int j = 0; j < l[i].size(); j++) {
    int t = l[i][j];
    coverRange(1, 0, n, loc[t], loc[t] + size[t], 1);
  }
  ans[i] = sum[1];
  for (int j = 0; j < ge[i].size(); j++) {
    int son = ge[i][j];
    if (v[son]) continue;
    solve(son, n);
  }
  for (int j = 0; j < l[i].size(); j++) {
    int t = l[i][j];
    coverRange(1, 0, n, loc[t], loc[t] + size[t], -1);
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    ge[a].push_back(b);
    ge[b].push_back(a);
  }
  memset(v, false, sizeof(v));
  memset(pre, -1, sizeof(pre));
  depth[0] = 0;
  dfs(0);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    int t = getLca(a, b);
    if (t == a || t == b) {
      l[t].push_back(t);
    } else {
      l[a].push_back(a);
      l[a].push_back(b);
      l[b].push_back(a);
      l[b].push_back(b);
    }
  }
  memset(v, false, sizeof(v));
  memset(cover, 0, sizeof(cover));
  memset(sum, 0, sizeof(sum));
  solve(0, n);
  for (int i = 0; i < n; i++) {
    int x = ans[i];
    if (x > 0) x--;
    if (i < n - 1) {
      printf("%d ", x);
    } else {
      printf("%d\n", x);
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
int cover[1 << 18], sum[1 << 18], ans[100000], loc[100000], size[100000],
    depth[100000], pre[100000][20];
vector<int> ge[100000], l[100000];
bool v[100000];
void dfs(int i) {
  static int cur = 0;
  loc[i] = cur++;
  size[i] = 1;
  v[i] = true;
  for (int j = 0; j < ge[i].size(); j++) {
    int son = ge[i][j];
    if (v[son]) continue;
    depth[son] = depth[i] + 1;
    pre[son][0] = i;
    for (int k = 1; 1 << k <= depth[son]; k++)
      pre[son][k] = pre[pre[son][k - 1]][k - 1];
    dfs(son);
    size[i] += size[son];
  }
}
int getLca(int a, int b) {
  if (depth[a] < depth[b]) swap(a, b);
  int lg = 0;
  while (1 << lg <= depth[a]) lg++;
  for (int i = lg - 1; i >= 0; i--)
    if (depth[a] - (1 << i) >= depth[b]) a = pre[a][i];
  if (a == b) return a;
  for (int i = lg - 1; i >= 0; i--)
    if (pre[a][i] != -1 && pre[a][i] != pre[b][i]) {
      a = pre[a][i];
      b = pre[b][i];
    }
  return pre[a][0];
}
void update(int i, int l, int r) {
  if (cover[i] == 0) {
    if (l + 1 == r) {
      sum[i] = 0;
    } else {
      sum[i] = sum[i * 2] + sum[i * 2 + 1];
    }
  } else {
    sum[i] = r - l;
  }
}
void coverRange(int i, int l, int r, int l0, int r0, int d) {
  if (l0 <= l && r <= r0) {
    cover[i] += d;
    update(i, l, r);
    return;
  }
  int mid = (l + r) / 2;
  if (l0 < mid) coverRange(i * 2, l, mid, l0, r0, d);
  if (r0 > mid) coverRange(i * 2 + 1, mid, r, l0, r0, d);
  update(i, l, r);
}
void solve(int i, int n) {
  v[i] = true;
  for (int j = 0; j < l[i].size(); j++) {
    int t = l[i][j];
    coverRange(1, 0, n, loc[t], loc[t] + size[t], 1);
  }
  ans[i] = sum[1];
  for (int j = 0; j < ge[i].size(); j++) {
    int son = ge[i][j];
    if (v[son]) continue;
    solve(son, n);
  }
  for (int j = 0; j < l[i].size(); j++) {
    int t = l[i][j];
    coverRange(1, 0, n, loc[t], loc[t] + size[t], -1);
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    ge[a].push_back(b);
    ge[b].push_back(a);
  }
  memset(v, false, sizeof(v));
  memset(pre, -1, sizeof(pre));
  depth[0] = 0;
  dfs(0);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    int t = getLca(a, b);
    if (t == a || t == b) {
      l[t].push_back(t);
    } else {
      l[a].push_back(a);
      l[a].push_back(b);
      l[b].push_back(a);
      l[b].push_back(b);
    }
  }
  memset(v, false, sizeof(v));
  memset(cover, 0, sizeof(cover));
  memset(sum, 0, sizeof(sum));
  solve(0, n);
  for (int i = 0; i < n; i++) {
    int x = ans[i];
    if (x > 0) x--;
    if (i < n - 1) {
      printf("%d ", x);
    } else {
      printf("%d\n", x);
    }
  }
  return 0;
}

```