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 + 7;
vector<int> ad[N];
int h[N], n, m, dp[N][30];
int d[N], R = 320;
vector<int> red;
void dfs(int v, int p) {
  h[v] = h[p] + 1;
  dp[v][1] = p;
  for (auto u : ad[v]) {
    if (u != p) dfs(u, v);
  }
}
int dis(int u, int v) {
  int res = 0;
  if (h[v] < h[u]) swap(v, u);
  for (int i = 20; i >= 1; i--) {
    if (h[v] - (1 << (i - 1)) >= h[u]) {
      res += (1 << (i - 1));
      v = dp[v][i];
    }
  }
  if (u == v) return res;
  for (int i = 20; i >= 1; i--) {
    if (dp[v][i] != dp[u][i]) {
      res += 2 * (1 << (i - 1));
      v = dp[v][i];
      u = dp[u][i];
    }
  }
  return res + 2;
}
void bfs() {
  deque<int> dq;
  memset(d, 0, sizeof d);
  for (auto u : red) dq.push_back(u), d[u] = 1;
  while (dq.size()) {
    int v = dq[0];
    dq.pop_front();
    for (auto u : ad[v]) {
      if (!d[u]) d[u] = d[v] + 1, dq.push_back(u);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    ad[u].push_back(v);
    ad[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) dp[i][0] = i;
  dfs(1, 0);
  for (int j = 2; j <= 20; j++)
    for (int i = 1; i <= n; i++) dp[i][j] = dp[dp[i][j - 1]][j - 1];
  vector<int> pos;
  red.push_back(1);
  bfs();
  while (m--) {
    int t;
    cin >> t;
    if (t == 1) {
      int tt;
      cin >> tt;
      pos.push_back(tt);
      if (pos.size() > R) {
        for (auto u : pos) {
          red.push_back(u);
        }
        bfs();
        pos.clear();
      }
    } else {
      int v;
      cin >> v;
      int res = d[v] - 1;
      for (auto u : pos) res = min(res, dis(v, u));
      cout << res << '\n';
    }
  }
  return 0;
}
