#include <bits/stdc++.h>
using namespace std;
void run();
int main() {
  ios::sync_with_stdio(0);
  run();
}
struct seg_t {
  seg_t *c[2];
  int n, sum;
  seg_t(int n, bool rec = true) : n(n), sum(0) {
    if (n == 1 or not rec) return;
    c[0] = c[1] = new seg_t(n / 2);
  }
  seg_t *insert(int l, int r) {
    if (sum == n or r <= 0 or l >= n) return this;
    seg_t *res;
    if (l <= 0 and r >= n) {
      if (sum == n) return this;
      res = new seg_t(n, false);
      res->sum = n;
    } else {
      seg_t *a = c[0]->insert(l, r);
      seg_t *b = c[1]->insert(l - n / 2, r - n / 2);
      if (a == c[0] and b == c[1]) return this;
      res = new seg_t(n, false);
      res->c[0] = a;
      res->c[1] = b;
      res->sum = a->sum + b->sum;
    }
    return res;
  }
};
vector<int> edge[100000];
vector<int> link[100000];
int res[1000000];
int lef[1000000];
int rgt[1000000];
void build(int i, int p, int x) {
  lef[i] = x;
  rgt[i] = x + 1;
  for (int j : edge[i]) {
    if (j != p) {
      build(j, i, rgt[i]);
      rgt[i] = rgt[j] + 1;
    }
  }
}
void dfs(int x, int p, seg_t *tree) {
  if (not link[x].empty()) {
    tree = tree->insert(lef[x], rgt[x] + 1);
  }
  for (int y : link[x]) {
    tree = tree->insert(lef[y], rgt[y] + 1);
  }
  res[x] = tree->sum;
  for (int y : edge[x]) {
    if (y != p) {
      dfs(y, x, tree);
    }
  }
}
void run() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    link[a].push_back(b);
    link[b].push_back(a);
  }
  build(0, -1, 0);
  seg_t root(1 << 18);
  dfs(0, -1, &root);
  for (int i = 0; i < n; i++) {
    cout << max(0, res[i] / 2 - 1) << (i + 1 == n ? '\n' : ' ');
  }
}

### Prompt

Your task is to create 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;
void run();
int main() {
  ios::sync_with_stdio(0);
  run();
}
struct seg_t {
  seg_t *c[2];
  int n, sum;
  seg_t(int n, bool rec = true) : n(n), sum(0) {
    if (n == 1 or not rec) return;
    c[0] = c[1] = new seg_t(n / 2);
  }
  seg_t *insert(int l, int r) {
    if (sum == n or r <= 0 or l >= n) return this;
    seg_t *res;
    if (l <= 0 and r >= n) {
      if (sum == n) return this;
      res = new seg_t(n, false);
      res->sum = n;
    } else {
      seg_t *a = c[0]->insert(l, r);
      seg_t *b = c[1]->insert(l - n / 2, r - n / 2);
      if (a == c[0] and b == c[1]) return this;
      res = new seg_t(n, false);
      res->c[0] = a;
      res->c[1] = b;
      res->sum = a->sum + b->sum;
    }
    return res;
  }
};
vector<int> edge[100000];
vector<int> link[100000];
int res[1000000];
int lef[1000000];
int rgt[1000000];
void build(int i, int p, int x) {
  lef[i] = x;
  rgt[i] = x + 1;
  for (int j : edge[i]) {
    if (j != p) {
      build(j, i, rgt[i]);
      rgt[i] = rgt[j] + 1;
    }
  }
}
void dfs(int x, int p, seg_t *tree) {
  if (not link[x].empty()) {
    tree = tree->insert(lef[x], rgt[x] + 1);
  }
  for (int y : link[x]) {
    tree = tree->insert(lef[y], rgt[y] + 1);
  }
  res[x] = tree->sum;
  for (int y : edge[x]) {
    if (y != p) {
      dfs(y, x, tree);
    }
  }
}
void run() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    link[a].push_back(b);
    link[b].push_back(a);
  }
  build(0, -1, 0);
  seg_t root(1 << 18);
  dfs(0, -1, &root);
  for (int i = 0; i < n; i++) {
    cout << max(0, res[i] / 2 - 1) << (i + 1 == n ? '\n' : ' ');
  }
}

```