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 MAX = 100005, SIZE = 300;
int n, m, dist[MAX], up[MAX][18], tin[MAX], tout[MAX], timer, depth[MAX];
vector<int> G[MAX];
void dfs(int u, int p, int dep) {
  tin[u] = ++timer;
  depth[u] = dep;
  up[u][0] = p;
  for (int i = 1; i < 18; i++) up[u][i] = up[up[u][i - 1]][i - 1];
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == p) continue;
    dfs(v, u, dep + 1);
  }
  tout[u] = ++timer;
}
bool isancestor(int u, int v) {
  return (tin[u] <= tin[v] && tout[v] <= tout[u]);
}
int lca(int u, int v) {
  if (isancestor(u, v)) return u;
  if (isancestor(v, u)) return v;
  if (depth[v] > depth[u]) swap(u, v);
  for (int i = 17; i >= 0; i--)
    if (!isancestor(up[u][i], v)) u = up[u][i];
  return up[u][0];
}
int getdist(int u, int v) { return depth[u] + depth[v] - 2 * depth[lca(u, v)]; }
void bfs(deque<int> Q) {
  for (int i = 0; i < Q.size(); i++) dist[Q[i]] = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop_front();
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (dist[v] > dist[u] + 1) {
        dist[v] = dist[u] + 1;
        Q.push_back(v);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  tin[0] = -123456789;
  tout[0] = 123456789;
  for (int i = 0; i <= n; i++) dist[i] = 123456789;
  dfs(1, 0, 0);
  bfs({1});
  deque<int> red;
  for (int i = 0; i < m; i++) {
    int op, u;
    cin >> op >> u;
    if (op == 1) {
      red.push_back(u);
      if (red.size() == SIZE) {
        bfs(red);
        red.clear();
      }
    } else {
      int res = dist[u];
      for (int j = 0; j < red.size(); j++) res = min(res, getdist(u, red[j]));
      cout << res << "\n";
    }
  }
}
