#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n;
vector<int> g[N];
int in[N], out[N], dfs_t;
void dfs_prec(int u, int p) {
  in[u] = ++dfs_t;
  for (int v : g[u])
    if (v != p) dfs_prec(v, u);
  out[u] = dfs_t;
}
int lz[4 * N];
pair<int, int> st[4 * N];
pair<int, int> comb(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first) return a;
  if (a.first > b.first) return b;
  return {a.first, a.second + b.second};
}
void build(int p, int L, int R) {
  st[p].second = R - L + 1;
  if (L == R) return;
  int mid = (L + R) / 2;
  build(2 * p, L, mid);
  build(2 * p + 1, mid + 1, R);
}
void prop(int p, int L, int R) {
  if (lz[p]) {
    st[p].first += lz[p];
    if (L < R) {
      lz[2 * p] += lz[p];
      lz[2 * p + 1] += lz[p];
    }
    lz[p] = 0;
  }
}
void upd(int p, int L, int R, int i, int j, int val) {
  prop(p, L, R);
  if (i > R or j < L) return;
  if (L >= i and R <= j) {
    lz[p] = val;
    prop(p, L, R);
    return;
  }
  int mid = (L + R) / 2;
  upd(2 * p, L, mid, i, j, val);
  upd(2 * p + 1, mid + 1, R, i, j, val);
  st[p] = comb(st[2 * p], st[2 * p + 1]);
}
vector<pair<int, int>> to_add[N];
int ans[N];
void dfs(int u, int p) {
  for (auto [l, r] : to_add[u]) {
    upd(1, 1, n, l, r, 1);
  }
  ans[u] = st[1].first ? n : n - st[1].second;
  for (int v : g[u])
    if (v != p) {
      dfs(v, u);
    }
  for (auto [l, r] : to_add[u]) {
    upd(1, 1, n, l, r, -1);
  }
}
int main() {
  int m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs_prec(1, 1);
  while (m--) {
    int a, b;
    scanf("%d %d", &a, &b);
    to_add[a].emplace_back(in[a], out[a]);
    to_add[a].emplace_back(in[b], out[b]);
    to_add[b].emplace_back(in[a], out[a]);
    to_add[b].emplace_back(in[b], out[b]);
  }
  build(1, 1, n);
  dfs(1, 1);
  for (int i = 1; i <= n; i++) {
    if (ans[i] > 0) ans[i]--;
    printf("%d%c", ans[i], " \n"[i == n]);
  }
}

### Prompt

Create a solution in CPP for 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 = 100005;
int n;
vector<int> g[N];
int in[N], out[N], dfs_t;
void dfs_prec(int u, int p) {
  in[u] = ++dfs_t;
  for (int v : g[u])
    if (v != p) dfs_prec(v, u);
  out[u] = dfs_t;
}
int lz[4 * N];
pair<int, int> st[4 * N];
pair<int, int> comb(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first) return a;
  if (a.first > b.first) return b;
  return {a.first, a.second + b.second};
}
void build(int p, int L, int R) {
  st[p].second = R - L + 1;
  if (L == R) return;
  int mid = (L + R) / 2;
  build(2 * p, L, mid);
  build(2 * p + 1, mid + 1, R);
}
void prop(int p, int L, int R) {
  if (lz[p]) {
    st[p].first += lz[p];
    if (L < R) {
      lz[2 * p] += lz[p];
      lz[2 * p + 1] += lz[p];
    }
    lz[p] = 0;
  }
}
void upd(int p, int L, int R, int i, int j, int val) {
  prop(p, L, R);
  if (i > R or j < L) return;
  if (L >= i and R <= j) {
    lz[p] = val;
    prop(p, L, R);
    return;
  }
  int mid = (L + R) / 2;
  upd(2 * p, L, mid, i, j, val);
  upd(2 * p + 1, mid + 1, R, i, j, val);
  st[p] = comb(st[2 * p], st[2 * p + 1]);
}
vector<pair<int, int>> to_add[N];
int ans[N];
void dfs(int u, int p) {
  for (auto [l, r] : to_add[u]) {
    upd(1, 1, n, l, r, 1);
  }
  ans[u] = st[1].first ? n : n - st[1].second;
  for (int v : g[u])
    if (v != p) {
      dfs(v, u);
    }
  for (auto [l, r] : to_add[u]) {
    upd(1, 1, n, l, r, -1);
  }
}
int main() {
  int m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs_prec(1, 1);
  while (m--) {
    int a, b;
    scanf("%d %d", &a, &b);
    to_add[a].emplace_back(in[a], out[a]);
    to_add[a].emplace_back(in[b], out[b]);
    to_add[b].emplace_back(in[a], out[a]);
    to_add[b].emplace_back(in[b], out[b]);
  }
  build(1, 1, n);
  dfs(1, 1);
  for (int i = 1; i <= n; i++) {
    if (ans[i] > 0) ans[i]--;
    printf("%d%c", ans[i], " \n"[i == n]);
  }
}

```