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;
long n, u, v, m, type;
long check[100005], child[100005], visit[100005];
long par_cen[100005], par[100005][20];
long h_cen[100005], h[100005], dist[100005];
vector<long> adj[100005];
vector<pair<long, long> > cen[100005];
void dfs(int u, int p) {
  child[u] = 1;
  for (int v : adj[u]) {
    if (v != p && check[v] == 0) {
      dfs(v, u);
      child[u] += child[v];
    }
  }
}
long findcen(int u, int p, int size) {
  for (int v : adj[u]) {
    if (v != p && check[v] == 0) {
      if (child[v] * 2 > size) {
        return findcen(v, u, size);
      }
    }
  }
  return u;
}
void centroid(int root, int p) {
  dfs(root, root);
  long cen = findcen(root, root, child[root]);
  check[cen] = 1;
  par_cen[cen] = p;
  for (int v : adj[cen]) {
    if (check[v] == 0) {
      centroid(v, cen);
    }
  }
}
void build(int u, int p) {
  visit[u] = 1;
  par[u][0] = p;
  h[u] = h[p] + 1;
  for (int v : adj[u]) {
    if (visit[v] == 0) {
      build(v, u);
    }
  }
}
long kc(int a, int b) {
  long u = a, v = b, lca;
  if (h[u] < h[v]) swap(u, v);
  for (int i = 19; i >= 0; i--) {
    if (h[u] - (1 << i) >= h[v]) {
      u = par[u][i];
    }
  }
  if (u == v) {
    lca = u;
  } else {
    for (int i = 19; i >= 0; i--) {
      if (par[u][i] != par[v][i] && par[u][i] != -1 && par[v][i] != -1) {
        u = par[u][i];
        v = par[v][i];
      }
    }
    lca = par[u][0];
  }
  return h[a] + h[b] - 2 * h[lca];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  centroid(1, 0);
  memset(par, -1, sizeof par);
  build(1, 0);
  for (int j = 1; (1 << j) < n; j++) {
    for (int i = 1; i <= n; i++) {
      if (par[i][j - 1] != -1) {
        par[i][j] = par[par[i][j - 1]][j - 1];
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    long tmp = i;
    while (tmp != 0) {
      cen[i].push_back(make_pair(tmp, kc(tmp, i)));
      tmp = par_cen[tmp];
    }
    dist[i] = 1e9;
  }
  for (int i = 0; i < cen[1].size(); i++) {
    dist[cen[1][i].first] = min(dist[cen[1][i].first], cen[1][i].second);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%ld %ld", &type, &v);
    if (type == 1) {
      for (int j = 0; j < cen[v].size(); j++) {
        dist[cen[v][j].first] = min(dist[cen[v][j].first], cen[v][j].second);
      }
    } else {
      long mn = 1e9;
      for (int j = 0; j < cen[v].size(); j++) {
        mn = min(mn, dist[cen[v][j].first] + cen[v][j].second);
      }
      printf("%ld\n", mn);
    }
  }
}
