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 long long mod = 1e9 + 7;
const int maxn = 1e5 + 5;
vector<int> adj[maxn];
const int lognum = 18;
int depth[100005], parent[100005][lognum], subtree_sum[100005];
void dfs(int v, int par) {
  parent[v][0] = par;
  depth[v] = (par == -1 ? 0 : depth[par] + 1);
  subtree_sum[v] = 1;
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int vv = adj[v][i];
    if (vv == par) {
      continue;
    }
    dfs(vv, v);
    subtree_sum[v] += subtree_sum[vv];
  }
}
int LCA(int a, int b) {
  if (depth[a] < depth[b]) swap(a, b);
  int diff = depth[a] - depth[b];
  for (int i = 0; i < lognum; i++) {
    if ((diff & (1 << i)) > 0) {
      a = parent[a][i];
    }
  }
  if (a == b) return a;
  for (int i = lognum - 1; i >= 0; i--) {
    if (parent[a][i] != parent[b][i]) {
      a = parent[a][i];
      b = parent[b][i];
    }
  }
  if (a == b) return a;
  a = parent[a][0];
  return a;
}
int distance(int a, int b) {
  return depth[a] + depth[b] - 2 * depth[LCA(a, b)];
}
vector<pair<int, int> > a;
bool color[maxn];
int d[maxn];
int n, m;
void g(int id, int par) {
  for (auto i : adj[id]) {
    if (i == par) continue;
    g(i, id);
    d[id] = min(d[id], 1 + d[i]);
  }
  if (color[id]) d[id] = 0;
}
void h(int id, int par, int val) {
  d[id] = min(d[id], val);
  for (auto i : adj[id]) {
    if (i == par) continue;
    h(i, id, d[id] + 1);
  }
}
void f() {
  for (int i = 1; i <= n; ++i) d[i] = 1e5 + 5;
  g(1, -1);
  h(1, -1, 1e5 + 5);
  vector<int> changingnow;
  for (auto i : a) {
    if (i.first == 1) {
      changingnow.emplace_back(i.second);
    } else {
      int curr = d[i.second];
      for (auto j : changingnow) {
        curr = min(curr, distance(i.second, j));
      }
      cout << curr << '\n';
      ;
    }
  }
  for (auto i : changingnow) {
    color[i] = 1;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  color[1] = 1;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    adj[x].emplace_back(y);
    adj[y].emplace_back(x);
  }
  memset(parent, -1, sizeof(parent));
  dfs(1, -1);
  for (int j = 0; j + 1 < lognum; j++) {
    for (int i = 1; i <= n; i++) {
      if (parent[i][j] != -1) {
        parent[i][j + 1] = parent[parent[i][j]][j];
      }
    }
  }
  int blen = sqrt(m);
  while (m > 0) {
    a.clear();
    for (int i = 0; i < blen; ++i) {
      if (m <= 0) break;
      m--;
      int t, vv;
      cin >> t >> vv;
      a.emplace_back(t, vv);
    }
    f();
  }
  return 0;
}
