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 fastMax(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y; }
int fastMin(int x, int y) { return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x; }
const long long int MAXN = 1e5 + 10;
const long long int LG = 20;
const long long int INF = 1e9;
set<long long int> graph[MAXN];
long long int par[MAXN], sub[MAXN], level[MAXN], ans[MAXN];
long long int dp[LG][MAXN];
long long int n, m;
void dfs0(long long int u) {
  for (long long int v : graph[u]) {
    if (v == dp[0][u]) continue;
    dp[0][v] = u;
    level[v] = level[u] + 1;
    dfs0(v);
  }
}
void preprocess() {
  level[1] = 0;
  dp[0][1] = 0;
  dfs0(1);
  for (long long int i = 1; i < LG; i++)
    for (long long int j = 1; j <= n; j++) dp[i][j] = dp[i - 1][dp[i - 1][j]];
}
long long int lca(long long int a, long long int b) {
  if (level[a] > level[b]) swap(a, b);
  long long int diff = level[b] - level[a];
  for (long long int i = 0; i < LG; i++)
    if (diff & (1 << i)) b = dp[i][b];
  if (a == b) return a;
  for (long long int i = LG - 1; i >= 0; i--) {
    if (dp[i][a] != dp[i][b]) a = dp[i][a], b = dp[i][b];
  }
  return dp[0][a];
}
long long int dist(long long int a, long long int b) {
  return level[a] + level[b] - 2 * level[lca(a, b)];
}
long long int nn;
void dfs1(long long int u, long long int p) {
  sub[u] = 1;
  nn++;
  for (long long int v : graph[u]) {
    if (v != p) {
      dfs1(v, u);
      sub[u] += sub[v];
    }
  }
}
long long int dfs2(long long int u, long long int p) {
  for (long long int v : graph[u]) {
    if (v != p && sub[v] > nn / 2) return dfs2(v, u);
  }
  return u;
}
void decompose(long long int root, long long int p) {
  nn = 0;
  dfs1(root, root);
  long long int centroid = dfs2(root, root);
  if (p == -1) p = centroid;
  par[centroid] = p;
  for (long long int v : graph[centroid]) {
    graph[v].erase(centroid);
    decompose(v, centroid);
  }
  graph[centroid].clear();
}
void update(long long int u) {
  long long int x = u;
  while (1) {
    ans[x] = min(ans[x], dist(x, u));
    if (x == par[x]) break;
    x = par[x];
  }
}
long long int query(long long int u) {
  long long int x = u;
  long long int ret = INF;
  while (1) {
    ret = min(ret, dist(x, u) + ans[x]);
    if (x == par[x]) break;
    x = par[x];
  }
  return ret;
}
void solve() {
  cin >> n >> m;
  for (int i = (1); i <= (n - 1); ++i) {
    long long int u, v;
    cin >> u >> v;
    graph[u].insert(v);
    graph[v].insert(u);
  }
  preprocess();
  decompose(1, -1);
  for (int i = (1); i <= (n); ++i) ans[i] = INF;
  update(1);
  while (m--) {
    long long int t, v;
    cin >> t >> v;
    if (t == 1)
      update(v);
    else
      cout << query(v) << endl;
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  t = 1;
  while (t--) solve();
}
