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 + 1, bin = 20;
int color[N], level[N], par[N][bin], dist[N];
vector<int> g[N];
void dfs(int cur, int p) {
  level[cur] = level[p] + 1;
  par[cur][0] = p;
  for (int child : g[cur]) {
    if (child == p) continue;
    dfs(child, cur);
  }
}
int get_par(int a, int h) {
  for (; h; h -= h & -h) a = par[a][(int)log2(h & -h)];
  return a;
}
int lca(int a, int b) {
  if (level[a] < level[b]) swap(a, b);
  a = get_par(a, level[a] - level[b]);
  if (a == b) return a;
  int i;
  for (i = bin - 1; i >= 0; i--) {
    if (par[a][i] != -1 and par[a][i] != par[b][i])
      a = par[a][i], b = par[b][i];
  }
  return par[a][0];
}
int distf(int a, int b) {
  int p = lca(a, b);
  return level[a] + level[b] - 2 * level[p];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, j, a, b, m;
  cin >> n >> m;
  for (i = 1; i <= n - 1; i++) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (i = 1; i <= n; i++)
    for (j = 0; j <= bin - 1; j++) par[i][j] = -1;
  level[0] = -1;
  dfs(1, 0);
  color[1] = 1;
  for (j = 1; j <= bin - 1; j++)
    for (i = 1; i <= n; i++) par[i][j] = par[par[i][j - 1]][j - 1];
  int t, x, q;
  vector<int> red;
  red.push_back(1);
  for (i = 1; i <= n; i++) dist[i] = 1e8;
  dist[1] = 0;
  int _i, _j;
  int block = sqrt(m);
  for (_i = 0; _i <= block + 1; _i++) {
    queue<int> q;
    for (int e : red) q.push(e);
    red.clear();
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      for (int child : g[cur]) {
        if (dist[child] > dist[cur] + 1) {
          q.push(child);
          dist[child] = dist[cur] + 1;
        }
      }
    }
    for (_j = 0; _j <= block - 1; _j++) {
      cin >> t >> x;
      if (t == 1) {
        if (color[x]) continue;
        color[x] = 1;
        dist[x] = 0;
        red.push_back(x);
      } else {
        int mn = dist[x];
        for (int e : red) mn = min(mn, distf(e, x));
        cout << mn << endl;
      }
      m--;
      if (m == 0) return 0;
    }
  }
  return 0;
}
