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 long long int maxn = 1e5 + 10, maxi = 4 * maxn, LOG = 25, INF = 3e18,
                    mod = 1e9 + 7, SQ = 333;
long long int n, m, h[maxn], par[maxn][LOG], dis[maxn];
vector<long long int> adj[maxn], blue;
void dfs(int u, int v = -1) {
  for (int i : adj[u])
    if (i != v) {
      h[i] = h[u] + 1, par[i][0] = u, dfs(i, u);
    }
}
inline void preprocess() {
  queue<int> q;
  for (int i : blue) dis[i] = 0, q.push(i);
  while (!q.empty()) {
    int fr = q.front();
    q.pop();
    for (int i : adj[fr])
      if (dis[i] > dis[fr] + 1) dis[i] = dis[fr] + 1, q.push(i);
  }
  return;
}
long long int lca(int u, int v) {
  if (u == v) return 0;
  if (h[u] < h[v]) swap(u, v);
  int node[2] = {u, v};
  int log = LOG - 1;
  while (log >= 0) {
    if (h[par[u][log]] >= h[v]) u = par[u][log];
    log--;
  }
  if (u != v) {
    log = LOG - 1;
    while (log >= 0) {
      if (par[u][log] != par[v][log]) u = par[u][log], v = par[v][log];
      log--;
    }
    u = par[u][0], v = par[v][0];
  }
  return (h[node[0]] - h[u]) + (h[node[1]] - h[u]);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  fill(dis, dis + maxn, INF);
  for (long long int i = 0; i < n - 1; i++) {
    long long int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  par[1][0] = 1;
  for (long long int j = 1; j < LOG; j++)
    for (long long int i = 1; i <= n; i++)
      par[i][j] = par[par[i][j - 1]][j - 1];
  blue.push_back(1);
  while (m--) {
    long long int type, node;
    cin >> type >> node;
    if (blue.size() >= SQ) {
      preprocess();
      blue.clear();
    }
    if (type == 1) {
      blue.push_back(node);
    } else {
      long long int ans = dis[node];
      for (long long int i : blue) {
        ans = min(ans, lca(node, i));
      }
      cout << ans << "\n" << flush;
    }
  }
  return 0;
}
