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 M = 1e5 + 10;
const int SQ = 320;
const int INF = 1 << 30;
const int L = 20;
vector<int> adj[M];
int ans[M];
int h[M];
int par[M][L];
int q[M];
int sz = 0;
int n, m;
void dfs(int v = 0, int p = -1) {
  for (int u : adj[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    par[u][0] = v;
    dfs(u, v);
  }
}
void bfs() {
  for (int i = 0; i < sz; i++) {
    int v = q[i];
    for (int u : adj[v]) {
      if (ans[v] + 1 < ans[u]) {
        ans[u] = ans[v] + 1;
        q[sz++] = u;
      }
    }
  }
  sz = 0;
}
int lca(int u, int v) {
  if (h[u] > h[v]) swap(u, v);
  int del = h[v] - h[u];
  for (int i = 0; i < L; i++) {
    if (del >> i & 1) v = par[v][i];
  }
  if (u == v) return u;
  for (int i = L - 1; i >= 0; i--) {
    if (par[v][i] != par[u][i]) {
      v = par[v][i];
      u = par[u][i];
    }
  }
  return par[v][0];
}
int dis(int u, int v) { return h[u] + h[v] - 2 * h[lca(u, v)]; }
int main() {
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    ans[i] = INF;
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < L; j++) {
      par[i][j] = -1;
    }
  }
  dfs();
  for (int i = 1; i < L; i++) {
    for (int v = 0; v < n; v++) {
      if (par[v][i - 1] == -1) continue;
      par[v][i] = par[par[v][i - 1]][i - 1];
    }
  }
  q[sz++] = 0;
  ans[0] = 0;
  for (int i = 0; i < m; i++) {
    if (i % SQ == 0) {
      bfs();
    }
    int t, v;
    cin >> t >> v;
    v--;
    if (t == 1) {
      ans[v] = 0;
      q[sz++] = v;
    } else {
      int tmp = ans[v];
      for (int i = 0; i < sz; i++) {
        int u = q[i];
        tmp = min(tmp, dis(v, u));
      }
      cout << tmp << endl;
    }
  }
  return 0;
}
