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;
vector<int> tree[100010];
int par[100010], sub[100010], level[100010], ans[100010];
int dp[20][100010];
int nn, m;
void dfs0(int u) {
  for (int i = 0; i < tree[u].size(); i++)
    if (dp[0][u] != tree[u][i]) {
      level[tree[u][i]] = level[u] + 1;
      dp[0][tree[u][i]] = u;
      dfs0(tree[u][i]);
    }
}
void preprocess() {
  level[1] = 0;
  dp[0][1] = 0;
  dfs0(1);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= nn; j++) dp[i][j] = dp[i - 1][dp[i - 1][j]];
}
int lca(int a, int b) {
  if (level[a] > level[b]) swap(a, b);
  int d = level[b] - level[a];
  for (int i = 0; i < 20; i++)
    if (d & (1 << i)) b = dp[i][b];
  if (a == b) return a;
  for (int i = 20 - 1; i >= 0; i--)
    if (dp[i][a] != dp[i][b]) a = dp[i][a], b = dp[i][b];
  return dp[0][a];
}
int dist(int u, int v) { return level[u] + level[v] - 2 * level[lca(u, v)]; }
int no;
int dfs1(int u, int par) {
  no++;
  sub[u] = 1;
  if (tree[u].size() == 1 && tree[u][0] == par) return sub[u];
  for (int i = 0; i < tree[u].size(); i++)
    if (tree[u][i] != par) sub[u] += dfs1(tree[u][i], u);
  return sub[u];
}
int dfs2(int u, int p) {
  for (int i = 0; i < tree[u].size(); i++)
    if (tree[u][i] != p && sub[tree[u][i]] > no / 2) return dfs2(tree[u][i], u);
  return u;
}
void decompose(int root, int p) {
  no = 0;
  dfs1(root, p);
  int cen = dfs2(root, p);
  par[cen] = p;
  for (int i = 0; i < tree[cen].size(); i++) {
    std::vector<int>::iterator position =
        std::find(tree[tree[cen][i]].begin(), tree[tree[cen][i]].end(), cen);
    tree[tree[cen][i]].erase(position);
    decompose(tree[cen][i], cen);
  }
  tree[cen].clear();
}
void update(int v) {
  int u = v;
  while (u) {
    ans[u] = min(ans[u], dist(u, v));
    u = par[u];
  }
}
int query(int v) {
  int rval = ans[v];
  int u = v;
  while (u) {
    u = par[u];
    rval = min(rval, ans[u] + dist(u, v));
  }
  return rval;
}
int main() {
  cin >> nn >> m;
  for (int i = 0; i < nn - 1; i++) {
    int a, b;
    cin >> a >> b;
    tree[a].push_back(b);
    tree[b].push_back(a);
  }
  preprocess();
  decompose(1, 0);
  for (int i = 1; i <= nn; i++) ans[i] = 1000000010;
  update(1);
  while (m--) {
    int t, v;
    cin >> t >> v;
    if (t == 1) update(v);
    if (t == 2) cout << query(v) << endl;
  }
  return 0;
}
