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;
vector<int> nums[100000];
int vertex[100000][17];
int now_ans[100000], in[100000], out[100000], depth[100000], t = 0;
vector<int> now_Red;
void bfs() {
  queue<pair<int, int> > q;
  for (int i = (0); i < (100000); i++) now_ans[i] = -1;
  for (int i : now_Red) {
    q.push({i, 0});
    now_ans[i] = 0;
  }
  int a, b;
  while (q.size()) {
    tie(a, b) = q.front();
    q.pop();
    for (int i : nums[a]) {
      if (now_ans[i] == -1) {
        now_ans[i] = b + 1;
        q.push({i, b + 1});
      }
    }
  }
}
void dfs(int n, int pa) {
  depth[n] = ~pa ? depth[pa] + 1 : 0;
  vertex[n][0] = pa;
  in[n] = t++;
  for (int i = (1); i < (17); i++) {
    int k = vertex[n][i - 1];
    if (k == -1) break;
    vertex[n][i] = vertex[k][i - 1];
  }
  for (int i : nums[n]) {
    if (i == pa) continue;
    dfs(i, n);
  }
  out[n] = t++;
}
int lca(int a, int b) {
  if (in[a] <= in[b] and out[a] >= out[b]) return a;
  if (in[b] <= in[a] and out[b] >= out[a]) return b;
  for (int i = 17 - 1; i >= 0; --i) {
    int k = vertex[a][i];
    if (k == -1) continue;
    if (!(in[k] <= in[b] and out[k] >= out[b])) a = k;
  }
  return vertex[a][0];
}
int find_length(int a, int b) {
  int k = lca(a, b);
  return depth[a] + depth[b] - 2 * depth[k];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q, a, b;
  cin >> n >> q;
  for (int i = (0); i < (n - 1); i++) {
    cin >> a >> b;
    a--;
    b--;
    nums[a].push_back(b);
    nums[b].push_back(a);
  }
  for (int i = (0); i < (100000); i++) {
    for (int j = (0); j < (17); j++) {
      vertex[i][j] = -1;
    }
  }
  dfs(0, -1);
  now_Red.push_back(0);
  bfs();
  while (q--) {
    cin >> a >> b;
    b--;
    if (a == 1) {
      now_Red.push_back(b);
      if (now_Red.size() % 350 == 0) {
        bfs();
      }
    } else {
      int ans = now_ans[b];
      for (int i = ((now_Red.size() / 350) * 350); i < (now_Red.size()); i++) {
        ans = min(find_length(now_Red[i], b), ans);
      }
      cout << ans << endl;
    }
  }
}
