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 const MAXN = 2e5 + 10, MAXK = 30, INF = 0x3f3f3f3f;
int n, m;
set<int> adj[MAXN];
int dp[MAXK][MAXN], prof[MAXN];
void DFS(int v) {
  set<int>::iterator it;
  for (it = adj[v].begin(); it != adj[v].end(); it++)
    if (*it != dp[0][v]) dp[0][*it] = v, prof[*it] = prof[v] + 1, DFS(*it);
}
void build() {
  dp[0][1] = 1;
  DFS(1);
  for (int k = 1; k < MAXK; k++)
    for (int i = 1; i <= n; i++) dp[k][i] = dp[k - 1][dp[k - 1][i]];
}
int lca(int x, int y) {
  if (prof[y] < prof[x]) swap(x, y);
  int dis = prof[y] - prof[x];
  for (int k = MAXK - 1; k >= 0; k--)
    if (dis & (1 << k)) y = dp[k][y];
  if (x == y) return x;
  for (int k = MAXK - 1; k >= 0; k--)
    if (dp[k][x] != dp[k][y]) x = dp[k][x], y = dp[k][y];
  return dp[0][x];
}
int size[MAXN], total, par[MAXN], ans[MAXN];
void dfs(int v, int pai) {
  set<int>::iterator it;
  size[v] = 1;
  total++;
  for (it = adj[v].begin(); it != adj[v].end(); it++)
    if (*it != pai) dfs(*it, v), size[v] += size[*it];
}
int find(int v, int pai) {
  set<int>::iterator it;
  for (it = adj[v].begin(); it != adj[v].end(); it++)
    if (*it != pai && size[*it] > total / 2) return find(*it, v);
  return v;
}
void divide_and_conquer(int v, int pai) {
  total = 0;
  dfs(v, -1);
  int center = find(v, -1);
  par[center] = pai;
  set<int>::iterator it;
  for (it = adj[center].begin(); it != adj[center].end(); it++) {
    adj[*it].erase(center);
    divide_and_conquer(*it, center);
  }
  adj[v].clear();
}
int dist(int u, int v) { return prof[u] + prof[v] - 2 * prof[lca(u, v)]; }
void update(int v) {
  int next = v;
  while (next != -1) {
    ans[next] = min(ans[next], dist(next, v));
    next = par[next];
  }
}
int query(int v) {
  int next = v, res = INF;
  while (next != -1) {
    res = min(res, ans[next] + dist(next, v));
    next = par[next];
  }
  return res;
}
int main() {
  cin.sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].insert(v);
    adj[v].insert(u);
  }
  build();
  divide_and_conquer(1, -1);
  for (int i = 1; i <= n; i++) ans[i] = INF;
  update(1);
  for (int i = 0; i < m; i++) {
    int t, v;
    cin >> t >> v;
    if (t == 1)
      update(v);
    else
      cout << query(v) << '\n';
  }
  return 0;
}
