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;
using ll = long long;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 1e5 + 5;
set<int> g[N];
int l = 18, timer, level[N];
int tout[N], tin[N], up[N][20];
int sub[N], centroidpar[N];
int nodes;
void lca_dfs(int v, int p) {
  tin[v] = ++timer;
  up[v][0] = p;
  for (int i = 1; i <= l; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
  for (int u : g[v]) {
    if (u != p) {
      level[u] = level[v] + 1;
      lca_dfs(u, v);
    }
  }
  tout[v] = ++timer;
}
bool is_ancestor(int u, int v) {
  return tin[u] <= tin[v] && tout[u] >= tout[v];
}
int lca(int u, int v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (int i = l; i >= 0; --i) {
    if (!is_ancestor(up[u][i], v)) u = up[u][i];
  }
  return up[u][0];
}
int distance(int u, int v) {
  return level[u] + level[v] - 2 * level[lca(u, v)];
}
void dfs(int k, int par) {
  nodes++;
  sub[k] = 1;
  for (auto it : g[k]) {
    if (it == par) continue;
    dfs(it, k);
    sub[k] += sub[it];
  }
}
int find_centroid(int k, int par) {
  for (auto it : g[k]) {
    if (it == par) continue;
    if (sub[it] > (nodes >> 1)) {
      return find_centroid(it, k);
    }
  }
  return k;
}
void decompose(int k, int par) {
  nodes = 0;
  dfs(k, k);
  int node = find_centroid(k, k);
  centroidpar[node] = par;
  for (auto it : g[node]) {
    g[it].erase(node);
    decompose(it, node);
  }
  g[node].clear();
}
int ans[N];
void update(int u) {
  int x = u;
  while (x != 0) {
    ans[x] = min(ans[x], distance(x, u));
    x = centroidpar[x];
  }
}
int query(int u) {
  int x = u;
  int ret = 1e9;
  while (x != 0) {
    ret = min(ret, distance(u, x) + ans[x]);
    x = centroidpar[x];
  }
  return ret;
}
int main() {
  io();
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].insert(v);
    g[v].insert(u);
  }
  lca_dfs(1, 1);
  decompose(1, 0);
  for (int i = 1; i <= n; i++) {
    ans[i] = 1e9;
  }
  update(1);
  while (q--) {
    int t, u;
    cin >> t >> u;
    if (t == 1) {
      update(u);
    } else {
      cout << query(u) << "\n";
    }
  }
}
