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;
int n, m;
vector<int> g[100001];
void read_graph() {
  int u, v;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
int centroid_parent[100001];
bool is_centroid[100001];
int dist[100001][18];
int level[100001];
int sz[100001];
void dfs(int nd, int p, int pc, int ldis) {
  sz[nd] = 1;
  if (pc != -1) {
    dist[nd][level[pc]] = ldis + 1;
  }
  for (int i = 0; i < g[nd].size(); i++) {
    int z = g[nd][i];
    if (!is_centroid[z] && z != p) {
      dfs(z, nd, pc, ldis + 1);
      sz[nd] += sz[z];
    }
  }
}
int find_centroid(int nd, int ts) {
  for (int i = 0; i < g[nd].size(); i++) {
    int z = g[nd][i];
    if (!is_centroid[z] && sz[z] < sz[nd] && sz[z] >= ts / 2)
      return find_centroid(z, ts);
  }
  return nd;
}
void get_centroid(int nd, int pc) {
  dfs(nd, -1, pc, 0);
  int u = find_centroid(nd, sz[nd]);
  centroid_parent[u] = pc;
  is_centroid[u] = true;
  if (pc == -1)
    level[u] = 0;
  else
    level[u] = level[pc] + 1;
  for (int i = 0; i < g[u].size(); i++) {
    int z = g[u][i];
    if (!is_centroid[z]) {
      get_centroid(z, u);
    }
  }
}
void centroid_decompose() {
  fill(is_centroid, is_centroid + 100001, false);
  get_centroid(1, -1);
}
int ans[100001];
void redify(int u) {
  int tmp = u;
  ans[tmp] = 0;
  while (1) {
    tmp = centroid_parent[tmp];
    if (tmp == -1) break;
    ans[tmp] = min(ans[tmp], dist[u][level[tmp]]);
  }
}
int solve(int u) {
  int tmp = u;
  long long res = ans[u];
  while (1) {
    tmp = centroid_parent[tmp];
    if (tmp == -1) break;
    res = min(res, dist[u][level[tmp]] + (long long)ans[tmp]);
  }
  return res;
}
void answer_query() {
  int u, v;
  while (m--) {
    scanf("%d %d", &u, &v);
    if (u == 1)
      redify(v);
    else
      printf("%d\n", solve(v));
  }
}
int main() {
  read_graph();
  centroid_decompose();
  fill(ans, ans + 100001, INT_MAX);
  redify(1);
  answer_query();
  return 0;
}
