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;
void Matrix() {
  while (true) {
    char t = '0' + rand() % 4;
    printf("%c", t > '1' ? ' ' : t);
  }
}
const int N = 100009;
int Q[N];
int dist[N];
vector<int> g[N];
int up[N][20];
int l;
int tip[N], v[N];
const int inf = 1e9;
int tin[N], tout[N];
int timer = 0;
int path[N];
vector<int> res;
void bfs() {
  int qh = 0, qt = 0;
  for (int i = 0; i < (int)res.size(); ++i) Q[qt++] = res[i], dist[res[i]] = 0;
  res.clear();
  while (qh < qt) {
    int v = Q[qh++];
    for (int i = 0; i < (int)g[v].size(); ++i) {
      int to = g[v][i];
      if (dist[to] > dist[v] + 1) dist[to] = dist[v] + 1, Q[qt++] = to;
    }
  }
}
void dfs(int v, int p) {
  tin[v] = ++timer;
  up[v][0] = p;
  for (int i = 1; i <= l; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
  for (int i = 0; i < (int)g[v].size(); ++i) {
    int to = g[v][i];
    if (to == p) continue;
    path[to] = path[v] + 1;
    dfs(to, v);
  }
  tout[v] = ++timer;
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = l; i >= 0; --i)
    if (!upper(up[a][i], b)) a = up[a][i];
  return up[a][0];
}
int get(int a, int b) { return path[a] + path[b] - 2 * path[lca(a, b)]; }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  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);
  }
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &tip[i], &v[i]);
  }
  for (int i = 1; i <= n; ++i) dist[i] = inf;
  l = 0;
  while ((1 << l) <= n) ++l;
  dfs(1, 1);
  res.push_back(1);
  bfs();
  int blocks = sqrt(m + .0) + 1;
  for (int i = 0; i < m; i += blocks) {
    for (int j = i; j < m && j < i + blocks; ++j) {
      if (tip[j] == 1) {
        res.push_back(v[j]);
      } else {
        int res = dist[v[j]];
        for (int k = i; k < j; k++)
          if (tip[k] == 1) res = min(res, get(v[j], v[k]));
        printf("%d\n", res);
      }
    }
    bfs();
  }
  return 0;
}
