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 long long MOD = 1e9 + 7;
const long long N = 2e5 + 5;
const long double pi = 3.14159265359;
const long long INF = (1LL << 63);
long long n, t, x, y, ty, from[N], vis[N], dep[N], ans[N];
vector<long long> v[N], lis;
vector<pair<long long, long long> > dis[N];
pair<long long, long long> p[N];
queue<pair<long long, long long> > que;
long long BFS(long long nd, long long pa) {
  que.push(make_pair(nd, pa));
  while ((long long)que.size()) {
    nd = que.front().first;
    pa = que.front().second;
    que.pop();
    for (long long i : v[nd]) {
      if (i == pa || vis[i]) continue;
      from[i] = nd;
      que.push(make_pair(i, nd));
    }
  }
  return nd;
}
void DFS(long long nd, long long pa, long long im) {
  for (long long i : v[nd]) {
    if (i == pa || vis[i]) continue;
    dep[i] = dep[nd] + 1;
    dis[i].push_back(make_pair(im, dep[i]));
    DFS(i, nd, im);
  }
}
void build(long long ea, long long pa) {
  long long na = BFS(ea, 0), nb = BFS(na, 0), tx;
  if (na == nb) {
    vis[ea] = 1;
    dis[ea].push_back(make_pair(ea, 0));
    p[na] = make_pair(pa, 1);
    return;
  }
  lis.clear();
  lis.push_back(tx = nb);
  do {
    lis.push_back(tx = from[tx]);
  } while (tx != na);
  tx = lis[(long long)lis.size() / 2];
  BFS(ea, 0);
  p[tx].first = pa;
  nb = tx;
  p[tx].second++;
  while (nb != ea) nb = from[nb], p[tx].second++;
  vis[tx] = 1;
  dep[tx] = 0;
  dis[tx].push_back(make_pair(tx, 0));
  DFS(tx, 0, tx);
  for (auto i : v[tx]) {
    if (vis[i]) continue;
    build(i, tx);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> t;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  build(1, 0);
  for (long long i = 1; i <= n; i++) ans[i] = 1e9;
  for (pair<long long, long long> i : dis[1])
    ans[i.first] = min(ans[i.first], i.second);
  while (t--) {
    cin >> ty >> x;
    if (ty == 1) {
      for (pair<long long, long long> i : dis[x]) {
        ans[i.first] = min(ans[i.first], i.second);
      }
    } else {
      long long tmp = 1e9;
      for (pair<long long, long long> i : dis[x]) {
        tmp = min(tmp, ans[i.first] + i.second);
      }
      cout << tmp << "\n";
    }
  }
  return 0;
}
