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, m, ll[101010], rr[101010], tmm, sg[404040][2], ans[101010];
vector<int> v[101010], vv[101010];
void dfsxu(int x, int y) {
  tmm++;
  ll[x] = tmm;
  for (int i = 0; i < v[x].size(); i++) {
    int z = v[x][i];
    if (y == z) {
      continue;
    }
    dfsxu(z, x);
  }
  rr[x] = tmm;
}
void build(int l, int r, int xh) {
  if (l == r) {
    sg[xh][0] = 1;
    return;
  }
  int mid = (l + r) / 2, xh1 = xh * 2, xh2 = xh * 2 + 1;
  build(l, mid, xh1);
  build(mid + 1, r, xh2);
  sg[xh][0] = sg[xh1][0] + sg[xh2][0];
}
void ggxx(int l, int r, int xh) {
  if (l == r) {
    if (sg[xh][1] > 0) {
      sg[xh][0] = 0;
    } else {
      sg[xh][0] = 1;
    }
  } else if (sg[xh][1] > 0) {
    sg[xh][0] = 0;
  } else {
    sg[xh][0] = sg[xh * 2][0] + sg[xh * 2 + 1][0];
  }
}
void gx(int l, int r, int ndl, int ndr, int xh, int xx) {
  if (l >= ndl && r <= ndr) {
    sg[xh][1] += xx;
    ggxx(l, r, xh);
    return;
  }
  int mid = (l + r) / 2, xh1 = xh * 2, xh2 = xh * 2 + 1;
  if (ndl <= mid) {
    gx(l, mid, ndl, ndr, xh1, xx);
  }
  if (ndr > mid) {
    gx(mid + 1, r, ndl, ndr, xh2, xx);
  }
  ggxx(l, r, xh);
}
void dfs(int x, int y) {
  if (vv[x].size() != 0) {
    gx(1, tmm, ll[x], rr[x], 1, 1);
  }
  for (int i = 0; i < vv[x].size(); i++) {
    int z = vv[x][i];
    gx(1, tmm, ll[z], rr[z], 1, 1);
  }
  ans[x] = max(0, n - 1 - sg[1][0]);
  for (int i = 0; i < v[x].size(); i++) {
    int z = v[x][i];
    if (z == y) {
      continue;
    }
    dfs(z, x);
  }
  if (vv[x].size() != 0) {
    gx(1, tmm, ll[x], rr[x], 1, -1);
  }
  for (int i = 0; i < vv[x].size(); i++) {
    int z = vv[x][i];
    gx(1, tmm, ll[z], rr[z], 1, -1);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfsxu(1, 0);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    vv[a].push_back(b);
    vv[b].push_back(a);
  }
  build(1, tmm, 1);
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;
  return 0;
}
