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;
vector<long long> gra[200005];
long long n, q;
set<long long> g[200005];
long long parD[200005];
vector<long long> CG[200005];
long long sub[200005];
long long level[200005];
long long vis[200005];
long long par[200005];
long long anse[200005];
long long DP[200005][22];
void dfs0(int u) {
  for (long long i = 0; i < gra[u].size(); ++i)
    if (gra[u][i] != DP[u][0]) {
      DP[gra[u][i]][0] = u;
      level[gra[u][i]] = level[u] + 1;
      dfs0(gra[u][i]);
    }
}
void preprocess() {
  level[0] = 0;
  DP[0][0] = 0;
  dfs0(1);
  for (long long i = 1; i < 21; i++)
    for (long long j = 1; j <= n; j++) DP[j][i] = DP[DP[j][i - 1]][i - 1];
}
long long lca(long long a, long long b) {
  if (level[a] > level[b]) swap(a, b);
  long long d = level[b] - level[a];
  for (long long i = 0; i < 21; i++)
    if (d & (1 << i)) b = DP[b][i];
  if (a == b) return a;
  for (long long i = 21 - 1; i >= 0; i--)
    if (DP[a][i] != DP[b][i]) a = DP[a][i], b = DP[b][i];
  return DP[a][0];
}
long long dist(long long u, long long v) {
  return level[u] + level[v] - 2 * level[lca(u, v)];
}
void getgraph() {
  cin >> n;
  cin >> q;
  int a, b;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b;
    gra[a].push_back(b);
    gra[b].push_back(a);
    g[a].insert(b);
    g[b].insert(a);
  }
}
int nn;
void dfs1(int u, int p) {
  sub[u] = 1;
  nn++;
  for (auto it = g[u].begin(); it != g[u].end(); it++)
    if (*it != p) {
      dfs1(*it, u);
      sub[u] += sub[*it];
    }
}
int dfs2(int u, int p) {
  for (auto it = g[u].begin(); it != g[u].end(); it++)
    if (*it != p && sub[*it] > nn / 2) return dfs2(*it, u);
  return u;
}
void decompose(int root, int p) {
  nn = 0;
  dfs1(root, root);
  int centroid = dfs2(root, root);
  if (p == -1) p = centroid;
  parD[centroid] = p;
  if (p != centroid) {
    CG[p].push_back(centroid);
    CG[centroid].push_back(p);
  }
  for (auto it = g[centroid].begin(); it != g[centroid].end(); it++) {
    g[*it].erase(centroid);
    decompose(*it, centroid);
  }
  g[centroid].clear();
}
void upd(long long u) {
  long long x = u;
  while (1) {
    anse[x] = min(anse[x], dist(x, u));
    if (x == parD[x] || parD[x] <= 0) break;
    x = parD[x];
  }
}
long long qry(long long u) {
  long long x = u;
  long long mini = 1000000007;
  while (1) {
    mini = min(mini, dist(x, u) + anse[x]);
    if (x == parD[x] || parD[x] <= 0) break;
    x = parD[x];
  }
  return mini;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  getgraph();
  decompose(1, -1);
  preprocess();
  for (long long i = 1; i <= n; ++i) {
    anse[i] = 1000000007;
  }
  long long type, node;
  upd(1);
  while (q--) {
    cin >> type >> node;
    if (type == 1) {
      upd(node);
    } else {
      cout << qry(node) << "\n";
    }
  }
  return 0;
}
