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 N = (int)1e5 + 7;
const int L = (int)2e1 + 7;
const int MOD = (int)1e9 + 7;
int tin[N], tout[N], timer = 0, up[N][L], dist[N];
vector<vector<int>> g(N);
bool is_anc(int u, int v) { return (tin[u] <= tin[v] && tout[u] >= tout[v]); }
void magic(int v, int p) {
  up[v][0] = p;
  for (int i = 1; i < L; i++) up[v][i] = up[up[v][i - 1]][i - 1];
}
void dfs(int v = 1, int p = 1, int cur_dist = 0) {
  dist[v] = cur_dist;
  tin[v] = ++timer;
  magic(v, p);
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs(to, v, cur_dist + 1);
  }
  tout[v] = ++timer;
}
int lca(int u, int v) {
  if (is_anc(u, v)) return u;
  if (is_anc(v, u)) return v;
  for (int i = L - 1; i >= 0; i--)
    if (!is_anc(up[u][i], v)) u = up[u][i];
  return up[u][0];
}
int calc(int u, int v) { return dist[u] + dist[v] - dist[lca(u, v)] * 2; }
int n, m, block;
vector<int> ans, d(N, INT_MAX);
vector<vector<pair<int, int>>> q(N);
void analyze(int x) {
  vector<int> t;
  queue<int> r;
  for (auto e : q[x]) {
    if (e.first == 1) {
      t.push_back(e.second);
      d[e.second] = 0;
      r.push(e.second);
    }
    if (e.first == 2) {
      int v = e.second, cur = d[v];
      for (auto u : t) cur = min(cur, calc(u, v));
      ans.push_back(cur);
    }
  }
  while (!r.empty()) {
    int v = r.front();
    r.pop();
    for (auto to : g[v]) {
      if (d[to] > d[v] + 1) {
        d[to] = d[v] + 1;
        r.push(to);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs();
  block = sqrt(m);
  q[0].push_back(make_pair(1, 1));
  for (int i = 0; i < m; i++) {
    int type, v;
    cin >> type >> v;
    q[i / block].push_back(make_pair(type, v));
  }
  for (int i = 0; i < N; i++)
    if (q[i].size() > 0) analyze(i);
  for (auto x : ans) cout << x << '\n';
  return 0;
}
