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;
int const N = 100100;
int n, m, a, b, t, v, Root;
int Sparce[N][20], Depth[N], dist[N];
vector<int> Tree[N];
vector<int> RedNode;
void DFS(int u = 1, int p = -1) {
  Sparce[u][0] = p;
  for (auto i : Tree[u])
    if (i != p) {
      Depth[i] = Depth[u] + 1;
      DFS(i, u);
    }
}
void Build() {
  for (int k = 1; k <= 17; k++)
    for (int i = 1; i <= n; i++)
      if (Sparce[i][k - 1] + 1) Sparce[i][k] = Sparce[Sparce[i][k - 1]][k - 1];
}
int LCA(int u, int v) {
  if (Depth[u] < Depth[v]) swap(u, v);
  for (int k = 17; k >= 0; k--)
    if (Depth[u] - (1 << k) >= Depth[v]) u = Sparce[u][k];
  if (u == v) return u;
  for (int k = 17; k >= 0; k--)
    if (Sparce[u][k] != Sparce[v][k]) {
      u = Sparce[u][k];
      v = Sparce[v][k];
    }
  return Sparce[u][0];
}
int Distance(int u, int v) {
  return Depth[u] + Depth[v] - 2 * Depth[LCA(u, v)];
}
void BFS() {
  memset(dist, -1, sizeof dist);
  queue<int> q;
  for (auto i : RedNode) {
    dist[i] = 0;
    q.push(i);
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto i : Tree[u])
      if (dist[i] == -1) {
        dist[i] = dist[u] + 1;
        q.push(i);
      }
  }
}
int main() {
  memset(Sparce, -1, sizeof Sparce);
  RedNode.push_back(1);
  scanf("%d%d", &n, &m);
  Root = sqrt(m);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    Tree[a].push_back(b);
    Tree[b].push_back(a);
  }
  DFS();
  Build();
  vector<int> Cur;
  for (int i = 0; i < m; i++) {
    if (i % Root == 0) {
      while (Cur.size()) {
        RedNode.push_back(Cur.back());
        Cur.pop_back();
      }
      BFS();
    }
    scanf("%d%d", &t, &v);
    if (t == 1) {
      Cur.push_back(v);
    } else {
      int Ans = dist[v];
      for (auto i : Cur) Ans = min(Ans, Distance(v, i));
      printf("%d\n", Ans);
    }
  }
  return 0;
}
