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;
struct ndo_ {
  vector<long long> ady;
} T[100010];
const long long r_ = 320, lg2 = 20;
long long n, m, a, b, sig, op, node, ans[100010];
long long f[100010][lg2 + lg2], prof[100010];
vector<long long> redNodes, changeToRed;
bool use[100010];
queue<pair<long long, long long> > c;
void dfs(long long currentNode, long long lastNode, long long prof_) {
  f[currentNode][0] = lastNode;
  prof[currentNode] = prof_;
  for (long long i = 0; i < T[currentNode].ady.size(); i++) {
    if (lastNode == T[currentNode].ady[i]) continue;
    dfs(T[currentNode].ady[i], currentNode, prof_ + 1);
  }
  return;
}
void bfs() {
  for (long long it = 0; it <= n; it++) use[it] = false;
  for (long long it = 0; it < redNodes.size(); it++) {
    c.push({redNodes[it], 0});
    use[redNodes[it]] = true;
    ans[redNodes[it]] = 0LL;
  }
  while (!c.empty()) {
    pair<long long, long long> p = c.front();
    c.pop();
    for (long long it = 0; it < T[p.first].ady.size(); it++) {
      pair<long long, long long> h = {T[p.first].ady[it], p.second + 1};
      if (!use[h.first]) {
        use[h.first] = true;
        ans[h.first] = h.second;
        c.push(h);
      }
    }
  }
  return;
}
long long lca(long long u, long long v) {
  if (prof[u] > prof[v]) swap(u, v);
  for (long long pot = lg2; pot >= 0; pot--)
    if ((prof[v] - (1LL << pot)) >= prof[u]) v = f[v][pot];
  if (u == v) return u;
  for (long long pot = lg2; pot >= 0; pot--) {
    if (f[u][pot] != f[v][pot]) {
      u = f[u][pot];
      v = f[v][pot];
    }
  }
  return f[u][0];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 1; i < n; i++) {
    cin >> a >> b;
    T[a].ady.push_back(b);
    T[b].ady.push_back(a);
  }
  dfs(1, 1, 0);
  redNodes.push_back(1);
  for (long long pot = 1; pot <= lg2; pot++)
    for (long long currentNode = 1; currentNode <= n; currentNode++)
      f[currentNode][pot] = f[f[currentNode][pot - 1]][pot - 1];
  for (long long i = 0; i < m; i++) {
    if (i == sig) {
      for (long long j = 0; j < changeToRed.size(); j++)
        redNodes.push_back(changeToRed[j]);
      changeToRed.clear();
      bfs();
      sig += r_;
    }
    cin >> op >> node;
    if (op == 1) {
      changeToRed.push_back(node);
    } else {
      long long currentAns = ans[node];
      for (long long j = 0; j < changeToRed.size(); j++) {
        long long myLCA = lca(node, changeToRed[j]);
        currentAns = min(currentAns, prof[node] + prof[changeToRed[j]] -
                                         prof[myLCA] - prof[myLCA]);
      }
      cout << currentAns << "\n";
    }
  }
  return 0;
}
