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 N = 1e5 + 7;
int n, m, par[N][20], h[N];
vector<int> g[N];
void dfs1(int u, int p) {
  h[u] = h[p] + 1;
  par[u][0] = p;
  for (int v : g[u])
    if (v != p) dfs1(v, u);
}
void dp() {
  for (int j = 1; j <= 19; j++)
    for (int i = 1; i <= n; i++) par[i][j] = par[par[i][j - 1]][j - 1];
}
int lca(int u, int v) {
  if (h[u] > h[v]) swap(u, v);
  int delta = h[v] - h[u];
  for (int j = 19; j >= 0; j--)
    if (delta & (1 << j)) v = par[v][j];
  if (u == v) return u;
  for (int j = 19; j >= 0; j--)
    if (par[u][j] != par[v][j]) u = par[u][j], v = par[v][j];
  return par[u][0];
}
int dist(int u, int v) {
  int p = lca(u, v);
  return h[u] + h[v] - 2 * h[p];
}
const int oo = 1e9 + 7;
int nchild[N], cdad[N], best[N];
bool cen[N];
void dfs2(int u, int p) {
  nchild[u] = 1;
  for (int v : g[u]) {
    if (v == p || cen[v]) continue;
    dfs2(v, u);
    nchild[u] += nchild[v];
  }
}
int get_center_x(int u, int p, int cnt) {
  for (int v : g[u]) {
    if (v == p || cen[v]) continue;
    if (nchild[v] > cnt / 2) return get_center_x(v, u, cnt);
  }
  return u;
}
void decomp(int u, int p) {
  dfs2(u, u);
  u = get_center_x(u, u, nchild[u]);
  cdad[u] = p;
  cen[u] = 1;
  for (int v : g[u]) {
    if (v == p || cen[v]) continue;
    decomp(v, u);
  }
}
void upd(int u) {
  best[u] = 0;
  int v = cdad[u];
  while (v) {
    best[v] = min(best[v], dist(u, v));
    v = cdad[v];
  }
  return;
}
int qr(int u) {
  int res = best[u];
  int v = cdad[u];
  while (v) {
    res = min(res, dist(u, v) + best[v]);
    v = cdad[v];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs1(1, 0);
  dp();
  decomp(1, 0);
  for (int i = 1; i <= n; i++) best[i] = oo;
  upd(1);
  for (int i = 1; i <= m; i++) {
    int k, v;
    cin >> k >> v;
    if (k == 1)
      upd(v);
    else
      cout << qr(v) << '\n';
  }
}
