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 = 1e5 + 10;
std::vector<vector<int> > adj_list;
vector<int> pending;
int lca[N][23];
int ans[N];
int lvl[N];
int n, m;
void dfs(int u, int par) {
  lvl[u] = lvl[par] + 1;
  ans[u] = ans[par] + 1;
  if (u == 1) ans[u] = 0;
  lca[u][0] = par;
  for (int i = 1; (1 << i) <= n; i++) {
    lca[u][i] = lca[lca[u][i - 1]][i - 1];
  }
  for (int i = 0; i < adj_list[u].size(); i++) {
    int v = adj_list[u][i];
    if (v != par) {
      dfs(v, u);
    }
  }
}
int Lca(int x, int y) {
  if (lvl[x] < lvl[y]) swap(x, y);
  int lg;
  for (lg = 0; (1 << lg) <= n; lg++)
    ;
  lg--;
  for (int i = lg; i >= 0; i--) {
    if (lvl[lca[x][i]] >= lvl[y]) x = lca[x][i];
  }
  if (x == y) return x;
  for (int i = lg; i >= 0; i--) {
    if (lca[x][i] != lca[y][i]) {
      x = lca[x][i];
      y = lca[y][i];
    }
  }
  return lca[x][0];
}
int dis(int x, int y) { return lvl[x] + lvl[y] - 2 * lvl[Lca(x, y)]; }
void bfs() {
  queue<int> q;
  for (int i = 0; i < pending.size(); i++)
    q.push(pending[i]), ans[pending[i]] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < adj_list[u].size(); i++) {
      int v = adj_list[u][i];
      if (ans[v] > ans[u] + 1) {
        ans[v] = ans[u] + 1;
        q.push(v);
      }
    }
  }
}
int main() {
  cin >> n >> m;
  adj_list.assign(n + 1, vector<int>());
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    adj_list[x].push_back(y);
    adj_list[y].push_back(x);
  }
  dfs(1, 0);
  int sqm = sqrt(m);
  for (int i = 1; i <= m; i++) {
    int cmd, pos;
    cin >> cmd >> pos;
    if (cmd & 1) {
      pending.push_back(pos);
    } else {
      int res = ans[pos];
      for (int i = 0; i < pending.size(); i++)
        res = min(res, dis(pos, pending[i]));
      cout << res << '\n';
    }
    if (i % sqm == 0) {
      bfs();
      pending.clear();
    }
  }
}
