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 INF = 2e9;
const int N = 300101;
int n, m, q, d[N], dist[N], up[20][N], tin[N], tout[N], timer = 0;
vector<int> g[N], red;
void bfs() {
  queue<int> q;
  for (int i = 1; i <= n; i++) d[i] = INF;
  for (int i = 0; i < int(red.size()); i++) {
    d[red[i]] = 0;
    q.push(red[i]);
  }
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < int(g[v].size()); i++)
      if (d[g[v][i]] > d[v] + 1) {
        d[g[v][i]] = d[v] + 1;
        q.push(g[v][i]);
      }
  }
}
bool mark[N];
void dfs(int v, int p = 1) {
  tin[v] = timer++;
  mark[v] = 1;
  up[0][v] = p;
  for (int i = 1; i < 20; i++) up[i][v] = up[i - 1][up[i - 1][v]];
  for (int i = 0; i < int(g[v].size()); i++)
    if (!mark[g[v][i]]) {
      dist[g[v][i]] = dist[v] + 1;
      dfs(g[v][i], v);
    }
  tout[v] = timer++;
}
bool get(int a, int b) { return (tin[b] >= tin[a] && tin[b] <= tout[a]); }
int lca(int a, int b) {
  if (get(a, b)) return a;
  if (get(b, a)) return b;
  for (int i = 19; i >= 0; i--)
    if (!get(up[i][a], b)) a = up[i][a];
  return up[0][a];
}
int rast(int a, int b) {
  int c = lca(a, b);
  return (dist[a] - dist[c] + dist[b] - dist[c]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < int(n - 1); i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1);
  red.push_back(1);
  q = 1250;
  vector<int> rr;
  for (int i = 0; i < int(m); i++) {
    if (i % q == 0) {
      for (int j = 0; j < int(rr.size()); j++) red.push_back(rr[j]);
      rr.clear();
      bfs();
    }
    int t, v;
    scanf("%d %d", &t, &v);
    if (t == 1)
      rr.push_back(v);
    else {
      int mini = d[v];
      for (int j = 0; j < int(rr.size()); j++) mini = min(mini, rast(v, rr[j]));
      printf("%d\n", mini);
    }
  }
  return 0;
}
