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 
#include <bits/stdc++.h>
using namespace std;
int n, timeIn[100100], timeOut[100100], step, ans[100100];
int f[600500], cnt[600500];
vector<int> a[100100], query[100100];
void visit(int x, int par) {
  timeIn[x] = ++step;
  for (int i = 0; i < int(a[x].size()); i++)
    if (a[x][i] != par) visit(a[x][i], x);
  timeOut[x] = step;
}
void add(int nd, int l, int r, int x, int y, int val) {
  if (l == x && r == y) {
    f[nd] += val;
    if (f[nd])
      cnt[nd] = r - l + 1;
    else
      cnt[nd] = cnt[nd * 2] + cnt[nd * 2 + 1];
  } else {
    int m = (l + r) / 2;
    if (x <= m) add(nd * 2, l, m, x, min(y, m), val);
    if (m < y) add(nd * 2 + 1, m + 1, r, max(x, m + 1), y, val);
    if (f[nd])
      cnt[nd] = r - l + 1;
    else
      cnt[nd] = cnt[nd * 2] + cnt[nd * 2 + 1];
  }
}
void calc(int x, int par) {
  for (int i = 0; i < int(query[x].size()); i++) {
    int y = query[x][i];
    add(1, 1, n, timeIn[y], timeOut[y], 1);
  }
  ans[x] = cnt[1];
  for (int i = 0; i < int(a[x].size()); i++)
    if (a[x][i] != par) calc(a[x][i], x);
  for (int i = 0; i < int(query[x].size()); i++) {
    int y = query[x][i];
    add(1, 1, n, timeIn[y], timeOut[y], -1);
  }
}
int main() {
  int Q, x, y;
  cin >> n >> Q;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  visit(1, 0);
  while (Q--) {
    cin >> x >> y;
    if (timeIn[x] > timeIn[y]) swap(x, y);
    if (timeOut[x] >= timeOut[y])
      query[x].push_back(x);
    else {
      query[x].push_back(y);
      query[x].push_back(x);
      query[y].push_back(x);
      query[y].push_back(y);
    }
  }
  calc(1, 0);
  for (int i = 1; i <= n; i++) cout << max(0, ans[i] - 1) << ' ';
}
