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> g[100100];
int d[100100], s, h, q[100100], n, m, a, b;
void bfs() {
  while (s != h) {
    int u = q[s++];
    for (int i = 0; i < g[u].size(); i++)
      if (d[g[u][i]] > d[u] + 1) d[g[u][i]] = d[u] + 1, q[h++] = g[u][i];
  }
  s = 0, h = 0;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n - 1; ++i)
    cin >> a >> b, g[a - 1].push_back(b - 1), g[b - 1].push_back(a - 1),
        d[a] = 1 << 30, d[b] = 1 << 30;
  q[h++] = 0, d[0] = 0, bfs();
  while (m--) {
    int t, v;
    cin >> t >> v;
    if (t == 1)
      d[v - 1] = 0, q[h++] = v - 1;
    else
      bfs(), cout << d[v - 1] << '\n';
  }
}
