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 = 100005;
int pa[N][18], h[N];
vector<int> g[N];
int n, m;
int red[N], cntRed = 1, near[N], buf[N], cntb = 0;
void dfs(int u) {
  for (int v : g[u]) {
    if (pa[u][0] != v) {
      pa[v][0] = u;
      h[v] = h[u] + 1;
      for (int i = 1; i < 18; ++i) {
        pa[v][i] = pa[pa[v][i - 1]][i - 1];
      }
      dfs(v);
    }
  }
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  int d = h[u] - h[v];
  for (int i = 0; i < 18; ++i) {
    if (d & (1 << i)) u = pa[u][i];
  }
  if (u == v) return u;
  for (int i = 17; i >= 0; --i) {
    if (pa[u][i] != pa[v][i]) {
      u = pa[u][i];
      v = pa[v][i];
    }
  }
  return pa[u][0];
}
int dis(int u, int v) { return h[u] + h[v] - 2 * h[lca(u, v)]; }
void init() {
  cin >> n >> m;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  red[++cntRed] = 1;
  dfs(1);
}
void bfs() {
  for (int i = 1; i <= n; ++i) near[i] = -1;
  queue<pair<int, int> > q;
  for (int i = 1; i <= cntRed; ++i) q.push({red[i], 0}), near[red[i]] = 0;
  while (!q.empty()) {
    pair<int, int> u = q.front();
    q.pop();
    for (int v : g[u.first]) {
      if (near[v] == -1) {
        near[v] = u.second + 1;
        q.push({v, near[v]});
      }
    }
  }
}
void solve() {
  int limit = (int)sqrt(m);
  int cur = limit;
  queue<pair<int, int> > q;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    q.push({x, y});
    if (i == cur || i == m) {
      bfs();
      while (!q.empty()) {
        int t = q.front().first;
        int v = q.front().second;
        q.pop();
        if (t == 1) {
          buf[++cntb] = v;
          red[++cntRed] = v;
        } else {
          int res = near[v];
          for (int i = 1; i <= cntb; ++i) {
            res = min(res, dis(v, buf[i]));
          }
          cout << res << '\n';
        }
      }
      cntb = 0;
      cur += limit;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  init();
  solve();
}
