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 maxn = 1e5 + 100, Log = 20;
int n, m, color[maxn], dp[maxn], dis[maxn];
int par[maxn][Log], dep[maxn];
vector<int> G[maxn], ac;
void Dep(int u, int fa, int dep) {
  dis[u] = dep;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    Dep(v, u, dep + 1);
  }
}
void LcaDfs(int u, int fa, int d) {
  dep[u] = d;
  if (u == 1) {
    for (int i = 0; i < Log; i++) par[u][i] = u;
  } else {
    par[u][0] = fa;
    for (int i = 1; i < Log; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  }
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    LcaDfs(v, u, d + 1);
  }
}
int up(int x, int step) {
  for (int i = 0; i < Log; i++)
    if (step & (1 << i)) x = par[x][i];
  return x;
}
int Lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  u = up(u, dep[u] - dep[v]);
  if (u == v) return u;
  for (int i = Log - 1; i >= 0; i--)
    if (par[u][i] != par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  return par[u][0];
}
void bfs() {
  queue<int> Q;
  for (int i = 0; i < ac.size(); i++) {
    Q.push(ac[i]);
    dp[ac[i]] = 0;
  }
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (dp[v] > dp[u] + 1) {
        dp[v] = dp[u] + 1;
        Q.push(v);
      }
    }
  }
  ac.clear();
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) dp[i] = maxn * 10;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  ac.push_back(1);
  bfs();
  Dep(1, 0, 0);
  LcaDfs(1, 0, 1);
  int block = sqrt(m);
  while (m--) {
    int op, a;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &a);
      ac.push_back(a);
      if (ac.size() == block) bfs();
    } else {
      scanf("%d", &a);
      int ans = dp[a];
      for (int i = 0; i < ac.size(); i++) {
        ans = min(ans, dis[a] + dis[ac[i]] - 2 * dis[Lca(a, ac[i])]);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
