Xenia the programmer has a tree consisting of n nodes. We will consider the tree nodes indexed from 1 to n. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.

The distance between two tree nodes v and u is the number of edges in the shortest path between v and u.

Xenia needs to learn how to quickly execute queries of two types:

  1. paint a specified blue node in red; 
  2. calculate which red node is the closest to the given one and print the shortest distance to the closest red node. 



Your task is to write a program which will execute the described queries.

Input

The first line contains two integers n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of nodes in the tree and the number of queries. Next n - 1 lines contain the tree edges, the i-th line contains a pair of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — an edge of the tree.

Next m lines contain queries. Each query is specified as a pair of integers ti, vi (1 ≤ ti ≤ 2, 1 ≤ vi ≤ n). If ti = 1, then as a reply to the query we need to paint a blue node vi in red. If ti = 2, then we should reply to the query by printing the shortest distance from some red node to node vi.

It is guaranteed that the given graph is a tree and that all queries are correct.

Output

For each second type query print the reply in a single line.

Examples

Input

5 4
1 2
2 3
2 4
4 5
2 1
2 5
1 2
2 5


Output

0
3
2
#include <bits/stdc++.h>
using namespace std;
const int M = 1e5 + 10;
deque<int> tot;
bool mark[M];
int dis[M];
vector<int> vec[M];
const int maxL = 18;
int lca[M][maxL];
int h[M];
int t[M];
int n;
void dfs(int v, int par = 0) {
  t[v] = 1;
  lca[v][0] = par;
  for (int i = 1; i < maxL; i++) lca[v][i] = lca[lca[v][i - 1]][i - 1];
  for (int u : vec[v])
    if (u != par) h[u] = h[v] + 1, dfs(u, v), t[v] += t[u];
}
int get_a(int v, int k) {
  for (int i = 0; i < maxL; i++)
    if ((k >> i) & 1) v = lca[v][i];
  return v;
}
int LCA(int v, int u) {
  if (h[v] < h[u]) swap(v, u);
  v = get_a(v, h[v] - h[u]);
  if (v == u) return v;
  for (int i = maxL - 1; i >= 0; i--)
    if (lca[v][i] != lca[u][i]) v = lca[v][i], u = lca[u][i];
  return lca[v][0];
}
void update(deque<int> q) {
  while (!q.empty()) {
    int v = q.front();
    q.pop_front();
    for (int u : vec[v]) {
      if (dis[u] > (dis[v] + 1)) dis[u] = dis[v] + 1, q.push_back(u);
    }
  }
}
int getdis(int v, int u) {
  int L = LCA(v, u);
  return h[v] + h[u] - 2 * h[L];
}
int main() {
  memset(dis, 31, sizeof dis);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  int sq = sqrt(n);
  dfs(1);
  tot.push_back(1);
  dis[1] = 0;
  update(tot);
  vector<int> ver;
  while (m--) {
    int t, v;
    cin >> t >> v;
    if (t == 1)
      ver.push_back(v), tot.push_back(v), dis[v] = 0;
    else {
      int ans = dis[v];
      for (int i : ver) ans = min(ans, getdis(v, i));
      cout << ans << endl;
    }
    if (ver.size() == sq) {
      update(tot);
      ver.clear();
    }
  }
}
