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 inf = 1e9;
const int N = 1e5 + 5;
int n, q, id, best[N], siz[N], del[N];
vector<int> adj[N], trees[N];
map<int, int> dep[N];
int dfs(int u, int p = 0) {
  siz[u] = 1;
  for (int v : adj[u]) {
    if (v != p && !del[v]) siz[u] += dfs(v, u);
  }
  return siz[u];
}
void dfs2(int u, int p = 0, int d = 0) {
  trees[u].push_back(id);
  dep[id][u] = d;
  for (int v : adj[u]) {
    if (v != p && !del[v]) dfs2(v, u, d + 1);
  }
}
int centroid(int u, int size, int p = 0) {
  for (int v : adj[u]) {
    if (v != p && !del[v] && 2 * siz[v] > size) return centroid(v, size, u);
  }
  return u;
}
void decompose(int u = 1) {
  int c = centroid(u, dfs(u));
  dfs2(c);
  for (int u : trees[1]) {
    if (u == id) {
      best[id] = dep[id][1];
      break;
    }
  }
  id++;
  del[c] = 1;
  for (int v : adj[c]) {
    if (!del[v]) decompose(v);
  }
}
int main() {
  cin >> n >> q;
  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 = 0; i < N; i++) best[i] = inf;
  decompose();
  for (int i = 0; i < q; i++) {
    int t, v;
    cin >> t >> v;
    if (t == 1) {
      for (int x : trees[v]) {
        if (dep[x][v] < best[x]) best[x] = dep[x][v];
      }
    } else {
      int ans = inf;
      for (int x : trees[v]) ans = min(ans, best[x] + dep[x][v]);
      cout << ans << '\n';
    }
  }
}
