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 int N = 200005;
int sz[N], par[N], n, k, m, u, v, ans[N], x;
set<int> V[N];
map<int, int> d[N];
void dfs(int u, int p) {
  sz[u] = 1;
  for (set<int>::iterator it = V[u].begin(); it != V[u].end(); it++) {
    if (*it != p) {
      dfs(*it, u);
      sz[u] += sz[*it];
    }
  }
}
int centroid(int u, int p, int n) {
  int mx = 0;
  for (set<int>::iterator it = V[u].begin(); it != V[u].end(); it++) {
    if (*it != p)
      if (sz[*it] > n / 2) return centroid(*it, u, n);
  }
  return u;
}
void dfs1(int u, int p, int c) {
  for (set<int>::iterator it = V[u].begin(); it != V[u].end(); it++) {
    if (*it != p) {
      d[c][*it] = d[c][u] + 1;
      dfs1(*it, u, c);
    }
  }
}
void newtree(int u, int p) {
  dfs(u, p);
  int c = centroid(u, p, sz[u]);
  dfs1(c, p, c);
  par[c] = p;
  vector<int> h(V[c].begin(), V[c].end());
  for (int j = 0; j < h.size(); j++) {
    V[c].erase(h[j]);
    V[h[j]].erase(c);
    newtree(h[j], c);
  }
}
void update(int u) {
  int v = u;
  while (u != 0) {
    ans[u] = min(ans[u], d[u][v]);
    u = par[u];
  }
}
int getans(int u) {
  int v = u;
  int pas = 1e7;
  while (u != 0) {
    pas = min(pas, d[u][v] + ans[u]);
    u = par[u];
  }
  return pas;
}
int main() {
  cin >> n >> m;
  for (k = 2; k <= n; k++) {
    cin >> u >> v;
    ans[k] = 1e7;
    V[u].insert(v);
    V[v].insert(u);
  }
  ans[1] = 1e7;
  newtree(1, 0);
  update(1);
  while (m--) {
    cin >> x >> u;
    if (x == 1) {
      update(u);
    } else
      cout << getans(u) << endl;
  }
}
