#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int t[4 * N];
int m[4 * N];
int d[4 * N];
void upd(int v, int x) {
  m[v] += x;
  d[v] += x;
}
void pull(int v) {
  m[v] = min(m[2 * v + 1], m[2 * v + 2]);
  t[v] = (m[2 * v + 1] == m[v] ? t[2 * v + 1] : 0) +
         (m[2 * v + 2] == m[v] ? t[2 * v + 2] : 0);
  d[v] = 0;
}
void push(int v) {
  if (d[v]) {
    upd(2 * v + 1, d[v]);
    upd(2 * v + 2, d[v]);
    d[v] = 0;
  }
}
void build(int v, int l, int r) {
  if (r - l == 1) {
    t[v] = 1;
    m[v] = 0;
    d[v] = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m, r);
  pull(v);
}
void upd(int v, int l, int r, int x, int tl, int tr) {
  if (r <= tl || tr <= l) {
    return;
  } else if (l <= tl && tr <= r) {
    return upd(v, x);
  } else {
    push(v);
    int tm = (tl + tr) / 2;
    upd(2 * v + 1, l, r, x, tl, tm);
    upd(2 * v + 2, l, r, x, tm, tr);
    pull(v);
  }
}
bool get(int v, int id, int l, int r) {
  if (r - l == 1) {
    return !!m[v];
  } else {
    push(v);
    int m = (l + r) / 2;
    return (id < m ? get(2 * v + 1, id, l, m) : get(2 * v + 2, id, m, r));
  }
}
vector<int> g[N];
vector<int> q[N];
int tin[N], tout[N];
int timer = 0;
void pre_dfs(int v, int p = -1) {
  tin[v] = timer++;
  for (int to : g[v]) {
    if (to != p) {
      pre_dfs(to, v);
    }
  }
  tout[v] = timer;
}
int c[N];
int n;
void dfs(int v, int p = -1) {
  if (!q[v].empty()) upd(0, tin[v], tout[v], 1, 0, n);
  for (int x : q[v]) upd(0, tin[x], tout[x], 1, 0, n);
  c[v] = (m[0] == 0 ? n - t[0] - get(0, tin[v], 0, n) : n - 1);
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v);
    }
  }
  if (!q[v].empty()) upd(0, tin[v], tout[v], -1, 0, n);
  for (int x : q[v]) upd(0, tin[x], tout[x], -1, 0, n);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    q[x].push_back(y);
    q[y].push_back(x);
  }
  build(0, 0, n);
  pre_dfs(0);
  dfs(0);
  for (int i = 0; i < n; i++) {
    cout << c[i] << " \n"[i == n - 1];
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 100005;
int t[4 * N];
int m[4 * N];
int d[4 * N];
void upd(int v, int x) {
  m[v] += x;
  d[v] += x;
}
void pull(int v) {
  m[v] = min(m[2 * v + 1], m[2 * v + 2]);
  t[v] = (m[2 * v + 1] == m[v] ? t[2 * v + 1] : 0) +
         (m[2 * v + 2] == m[v] ? t[2 * v + 2] : 0);
  d[v] = 0;
}
void push(int v) {
  if (d[v]) {
    upd(2 * v + 1, d[v]);
    upd(2 * v + 2, d[v]);
    d[v] = 0;
  }
}
void build(int v, int l, int r) {
  if (r - l == 1) {
    t[v] = 1;
    m[v] = 0;
    d[v] = 0;
    return;
  }
  int m = (l + r) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m, r);
  pull(v);
}
void upd(int v, int l, int r, int x, int tl, int tr) {
  if (r <= tl || tr <= l) {
    return;
  } else if (l <= tl && tr <= r) {
    return upd(v, x);
  } else {
    push(v);
    int tm = (tl + tr) / 2;
    upd(2 * v + 1, l, r, x, tl, tm);
    upd(2 * v + 2, l, r, x, tm, tr);
    pull(v);
  }
}
bool get(int v, int id, int l, int r) {
  if (r - l == 1) {
    return !!m[v];
  } else {
    push(v);
    int m = (l + r) / 2;
    return (id < m ? get(2 * v + 1, id, l, m) : get(2 * v + 2, id, m, r));
  }
}
vector<int> g[N];
vector<int> q[N];
int tin[N], tout[N];
int timer = 0;
void pre_dfs(int v, int p = -1) {
  tin[v] = timer++;
  for (int to : g[v]) {
    if (to != p) {
      pre_dfs(to, v);
    }
  }
  tout[v] = timer;
}
int c[N];
int n;
void dfs(int v, int p = -1) {
  if (!q[v].empty()) upd(0, tin[v], tout[v], 1, 0, n);
  for (int x : q[v]) upd(0, tin[x], tout[x], 1, 0, n);
  c[v] = (m[0] == 0 ? n - t[0] - get(0, tin[v], 0, n) : n - 1);
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v);
    }
  }
  if (!q[v].empty()) upd(0, tin[v], tout[v], -1, 0, n);
  for (int x : q[v]) upd(0, tin[x], tout[x], -1, 0, n);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    q[x].push_back(y);
    q[y].push_back(x);
  }
  build(0, 0, n);
  pre_dfs(0);
  dfs(0);
  for (int i = 0; i < n; i++) {
    cout << c[i] << " \n"[i == n - 1];
  }
  return 0;
}

```