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 MAX = 1e5 + 9, lg = 20, SQ = 350;
vector<int> g[MAX];
int par[MAX][lg], h[MAX], d[MAX], n, m;
vector<int> v;
void dfs(int v, int p = -1) {
  for (auto u : g[v])
    if (u != p) {
      d[u] = h[u] = h[v] + 1, par[u][0] = v;
      for (int i = 1; i < lg; i++) par[u][i] = par[par[u][i - 1]][i - 1];
      dfs(u, v);
    }
}
int parent(int v, int h) {
  for (int i = 0; i < lg; i++)
    if (h >> i & 1) v = par[v][i];
  return v;
}
int lca(int v, int u) {
  if (h[v] < h[u]) swap(v, u);
  v = parent(v, h[v] - h[u]);
  if (v == u) return v;
  for (int i = lg - 1; i >= 0; i--)
    if (par[v][i] != par[u][i]) v = par[v][i], u = par[u][i];
  return par[v][0];
}
int dis(int v, int u) { return h[v] + h[u] - 2 * h[lca(v, u)]; }
queue<int> q;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1, v, u; i < n; i++)
    cin >> v >> u, v--, u--, g[v].push_back(u), g[u].push_back(v);
  dfs(0);
  while (m--) {
    int t, u;
    cin >> t >> u, u--;
    if (t == 1)
      v.push_back(u);
    else {
      int ans = d[u];
      for (auto w : v) ans = min(ans, dis(u, w));
      cout << ans << '\n';
    }
    if (v.size() == SQ) {
      for (auto w : v) q.push(w), d[w] = 0;
      while (q.size()) {
        int v = q.front();
        for (auto u : g[v])
          if (d[u] > d[v] + 1) d[u] = d[v] + 1, q.push(u);
        q.pop();
      }
      v.clear();
    }
  }
}
