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 INFint = 2147483547;
const long long INF = 9223372036854775807ll;
const long long MOD = 1000000007ll;
const long double EPS = 1e-13;
vector<vector<int> > g;
int LG = 32;
vector<int> tin, tout, d, d1;
vector<vector<int> > up;
int timer = 0;
void dfs(int v, int len = 0, int p = 0) {
  tin[v] = ++timer;
  up[v][0] = p;
  d[v] = len;
  for (int i = 1; i < LG; i++) up[v][i] = up[up[v][i - 1]][i - 1];
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    dfs(to, len + 1, v);
  }
  tout[v] = ++timer;
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = LG - 1; i > -1; i--)
    if (!upper(up[a][i], b)) a = up[a][i];
  return up[a][0];
}
int dist(int a, int b) { return d[a] + d[b] - 2 * d[lca(a, b)]; }
int n, m;
void bfs(vector<int> cur) {
  queue<int> q;
  for (int i = 0; i < cur.size(); i++) {
    d1[cur[i]] = 0;
    q.push(cur[i]);
  }
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < g[v].size(); i++) {
      int to = g[v][i];
      if (d1[v] + 1 < d1[to]) {
        d1[to] = d1[v] + 1;
        q.push(to);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  g.resize(n);
  d.resize(n);
  tin.resize(n);
  tout.resize(n);
  d.assign(n, INFint);
  d1.assign(n, INFint);
  up.resize(n, vector<int>(LG + 1));
  int bs = sqrt(m);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  vector<int> cur;
  cur.push_back(0);
  for (int i = 0; i < m; i++) {
    if (i % bs == 0) {
      bfs(cur);
      if (0) cerr << "masha\n";
      for (int v : cur) {
        if (0) cerr << v << ' ' << d1[v] << endl;
      }
      cur.clear();
    }
    int t, x;
    cin >> t >> x;
    x--;
    if (t == 1) {
      cur.push_back(x);
      continue;
    }
    int ans = d1[x];
    for (int j = 0; j < cur.size(); j++) {
      ans = min(ans, dist(cur[j], x));
    }
    cout << ans << endl;
  }
  fprintf(stderr, "\nTIME = %lf\n", 1.0 * clock() / CLOCKS_PER_SEC);
  ;
  return 0;
}
