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;
const int K = 20;
int F[N][K], level[N], d[N];
bool red[N];
vector<int> g[N];
void dfs(int x, int y = 0, int lvl = 0) {
  F[x][0] = y;
  for (int i = 1; i < K; i++) F[x][i] = F[F[x][i - 1]][i - 1];
  level[x] = lvl;
  for (auto v : g[x])
    if (v != y) {
      dfs(v, x, lvl + 1);
    }
}
void bfs(int n) {
  queue<pair<int, int>> q;
  for (int i = 0; i < n; i++)
    if (red[i]) q.push({i, 1});
  while (!q.empty()) {
    int x = q.front().first;
    int t = q.front().second;
    q.pop();
    if (d[x]) continue;
    d[x] = t;
    for (auto v : g[x]) q.push({v, t + 1});
  }
}
int lca(int x, int y) {
  if (level[x] > level[y]) swap(x, y);
  int d = level[y] - level[x];
  for (int i = 0; i < K; i++)
    if ((1 << i) & d) y = F[y][i];
  if (x == y) return x;
  for (int i = K - 1; i >= 0; i--) {
    if (F[x][i] != F[y][i]) x = F[x][i], y = F[y][i];
  }
  return F[x][0];
}
int dist(int x, int y) { return level[x] + level[y] - 2 * level[lca(x, y)]; }
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  red[0] = 1;
  int block = sqrt(m);
  int counter = 0;
  vector<int> s;
  while (m--) {
    if (!counter) {
      for (int i = 0; i < n; i++) d[i] = 0;
      s.clear();
      bfs(n);
      counter = block;
    }
    int t, v;
    cin >> t >> v;
    v--;
    if (t == 1) {
      red[v] = 1;
      s.push_back(v);
    } else {
      int r = d[v] - 1;
      for (auto x : s) r = min(r, dist(v, x));
      cout << r << '\n';
    }
    counter--;
  }
  return 0;
}
