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 N = 1e5 + 5;
long long color[N];
vector<long long> adj[N];
long long dis[N];
long long child[N];
vector<pair<long long, long long>> par[N];
bool vs[N];
void dfs(long long e, long long p) {
  child[e] = 1;
  for (auto x : adj[e]) {
    if (x != p && !vs[x]) {
      dfs(x, e);
      child[e] += child[x];
    }
  }
}
long long findcen(long long e, long long p, long long size) {
  for (auto x : adj[e]) {
    if (!vs[x] && x != p && child[x] > size / 2) {
      return findcen(x, e, size);
    }
  }
  return e;
}
void makepair(long long u, long long p, long long cen, long long d) {
  par[u].push_back({cen, d});
  for (auto x : adj[u]) {
    if (!vs[x] && x != p) {
      makepair(x, u, cen, d + 1);
    }
  }
}
void centroid(long long root) {
  dfs(root, root);
  long long cen = findcen(root, root, child[root]);
  makepair(cen, cen, cen, 0);
  vs[cen] = true;
  for (auto x : adj[cen]) {
    if (!vs[x]) {
      centroid(x);
    }
  }
}
void update(long long u) {
  for (auto x : par[u]) {
    dis[x.first] = min(dis[x.first], x.second);
  }
}
long long get(long long u) {
  long long ans = 1e9;
  for (auto x : par[u]) {
    ans = min(ans, dis[x.first] + x.second);
  }
  return ans;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  centroid(1);
  memset(dis, 69, sizeof(dis));
  update(1);
  for (int i = 1; i <= m; i++) {
    long long t, v;
    cin >> t >> v;
    if (t == 1) {
      update(v);
    } else if (t == 2) {
      cout << get(v) << endl;
    }
  }
}
