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 maxn = 1e5 + 100;
const int maxl = 20;
const int sqt = 300;
vector<int> adj[maxn];
vector<int> qu;
queue<int> q;
int par[maxn][maxl];
int dp[maxn];
int h[maxn];
bool mark[maxn];
void dfs(int v) {
  mark[v] = true;
  dp[v] = h[v];
  for (int i = 1; i < maxl and par[v][i - 1] != -1; i++)
    par[v][i] = par[par[v][i - 1]][i - 1];
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (!mark[u]) {
      h[u] = h[v] + 1;
      par[u][0] = v;
      dfs(u);
    }
  }
}
void bfs() {
  qu.clear();
  while (q.size()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < adj[v].size(); i++) {
      int u = adj[v][i];
      if (dp[v] + 1 < dp[u]) {
        dp[u] = dp[v] + 1;
        q.push(u);
      }
    }
  }
}
int lca(int v, int u) {
  if (h[v] < h[u]) swap(v, u);
  for (int i = maxl - 1; i >= 0; i--)
    if (par[v][i] != -1 and h[par[v][i]] >= h[u]) v = par[v][i];
  if (v == u) return v;
  for (int i = maxl - 1; i >= 0; i--)
    if (par[v][i] != -1 and par[v][i] != par[u][i])
      v = par[v][i], u = par[u][i];
  return par[v][0];
}
int get(int v, int u) { return h[u] + h[v] - 2 * h[lca(u, v)]; }
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  memset(par, -1, sizeof par);
  dfs(1);
  while (m--) {
    int t, v;
    cin >> t >> v;
    if (t == 1) {
      dp[v] = 0;
      q.push(v);
      qu.push_back(v);
      if (qu.size() > sqt) bfs();
    } else {
      int dis = dp[v];
      for (int i = 0; i < qu.size(); i++) dis = min(dis, get(qu[i], v));
      cout << dis << '\n';
    }
  }
  return 0;
}
