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 int N = 1e5 + 5;
const int LOG = 17;
int n, m;
vector<int> v[N];
vector<int> colored;
int col[N];
int D[N];
int par[N][LOG];
int deg[N];
vector<pair<int, int> > q[350];
void dfs(int s, int p) {
  if (p != -1)
    deg[s] = deg[p] + 1;
  else
    deg[s] = 1;
  par[s][0] = p;
  for (auto e : v[s]) {
    if (e != p) {
      dfs(e, s);
    }
  }
}
void BuildSparse() {
  memset(par, -1, sizeof(par));
  dfs(1, -1);
  for (int j = 1; j < LOG; j++) {
    for (int i = 1; i <= n; i++) {
      if (par[i][j - 1] != -1) par[i][j] = par[par[i][j - 1]][j - 1];
    }
  }
}
int LCA(int t, int s) {
  if (deg[t] > deg[s]) swap(t, s);
  int dif = deg[s] - deg[t];
  for (int i = 0; i < LOG; i++)
    if ((1 << i) & dif) s = par[s][i];
  if (t == s) return t;
  for (int i = LOG - 1; i >= 0; i--) {
    if (par[t][i] != par[s][i]) {
      t = par[t][i];
      s = par[s][i];
    }
  }
  return par[t][0];
}
int Dist(int i, int j) { return deg[i] + deg[j] - 2 * deg[LCA(i, j)]; }
void BFS() {
  queue<int> qq;
  for (int i = 1; i <= n; i++) D[i] = N;
  for (int i = 1; i <= n; i++)
    if (col[i]) {
      qq.push(i);
      D[i] = 0;
    }
  while (!qq.empty()) {
    int el = qq.front();
    qq.pop();
    for (auto e : v[el]) {
      if (D[el] + 1 < D[e]) {
        D[e] = D[el] + 1;
        qq.push(e);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  int x, y;
  for (int i = 1; i <= n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    q[i / 317].push_back({x, y});
  }
  BuildSparse();
  col[1] = 1;
  for (int i = 0; i <= m / 317; i++) {
    BFS();
    colored.clear();
    for (auto e : q[i]) {
      x = e.first;
      y = e.second;
      if (x == 1) {
        col[y] = 1;
        colored.push_back(y);
      } else {
        int sol = D[y];
        for (auto f : colored) {
          sol = min(sol, Dist(y, f));
        }
        cout << sol << "\n";
      }
    }
  }
  return 0;
}
