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;
queue<pair<int, int> > q;
int pa[18 + 5][200002], depth[200002], ans[200002];
vector<int> adj[200002], buffer;
void bfs() {
  while (!q.empty()) {
    int u = q.front().first;
    int d = q.front().second;
    q.pop();
    for (auto v : adj[u]) {
      if (d + 1 < ans[v]) {
        ans[v] = d + 1;
        q.push(make_pair(v, d + 1));
      }
    }
  }
}
void dfs(int u = 1, int p = -1) {
  pa[0][u] = p;
  for (auto v : adj[u]) {
    if (v == p) continue;
    depth[v] = depth[u] + 1;
    dfs(v, u);
  }
}
int LCA(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  for (int i = 18; i >= 0; --i) {
    if (depth[u] - (1 << i) >= depth[v]) {
      u = pa[i][u];
    }
  }
  if (u == v) return u;
  for (int i = 18; i >= 0; --i) {
    if (pa[i][u] != pa[i][v]) {
      u = pa[i][u];
      v = pa[i][v];
    }
  }
  return pa[0][u];
}
int dist(int u, int v) {
  int x = LCA(u, v);
  return depth[u] + depth[v] - (2LL) * depth[x];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i < n + 1; i++) ans[i] = 1000000006;
  ans[1] = 0;
  q.push(make_pair(1, 0));
  bfs();
  dfs();
  for (int i = 1; i <= 18; ++i) {
    for (int j = 1; j <= n; ++j) {
      pa[i][j] = pa[i - 1][pa[i - 1][j]];
    }
  }
  while (m--) {
    int t, v;
    cin >> t >> v;
    if (t == 1) {
      ans[v] = 0;
      buffer.push_back(v);
      q.push(make_pair(v, 0));
    } else {
      int res = ans[v];
      for (auto u : buffer) {
        res = min(res, dist(u, v));
      }
      cout << res << "\n";
    }
    if (buffer.size() == 80) {
      bfs();
      buffer.clear();
    }
  }
  return 0;
}
