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;
using ll = long long;
const int maxn = 1e5 + 5, inf = 1e9;
bool removed[maxn];
int n, q, sz[maxn], par[maxn], ans[maxn];
vector<int> vs[maxn], ct[maxn], dis[maxn];
int dfs_sz(int v, int p = -1) {
  sz[v] = 1;
  for (int e : vs[v])
    if (e != p && !removed[e]) sz[v] += dfs_sz(e, v);
  return sz[v];
}
int centroid(int v, int p, int tsz) {
  for (int e : vs[v])
    if (e != p && !removed[e] && (sz[e] << 1) > tsz) return centroid(e, v, tsz);
  return v;
}
void dfs_c(int v, int p, int d) {
  dis[v].emplace_back(d++);
  for (int e : vs[v])
    if (e != p && !removed[e]) dfs_c(e, v, d);
}
void decompose(int v, int p = -1) {
  int c = centroid(v, -1, dfs_sz(v));
  par[c] = p;
  if (p != -1) ct[p].emplace_back(c);
  removed[c] = true;
  for (int e : vs[c])
    if (!removed[e]) dfs_c(e, c, 1);
  for (int e : vs[c])
    if (!removed[e]) decompose(e, c);
}
void upd(int v) {
  for (int u = v, anc = 0; u != -1; u = par[u], ++anc)
    ans[u] = min(ans[u], dis[v][anc]);
}
int query(int v) {
  int res = inf;
  for (int u = v, anc = 0; u != -1; u = par[u], ++anc)
    res = min(res, ans[u] + dis[v][anc]);
  return res;
}
void init_cdt() {
  decompose(1);
  for (int i = 1; i <= n; i++) {
    ans[i] = inf;
    dis[i].emplace_back(0);
    reverse(dis[i].begin(), dis[i].end());
  }
  upd(1);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d %d", &x, &y);
    vs[x].emplace_back(y);
    vs[y].emplace_back(x);
  }
  init_cdt();
  for (int t, v; q--;) {
    scanf("%d %d", &t, &v);
    if (t == 1)
      upd(v);
    else
      printf("%d\n", query(v));
  }
  return 0;
}
